Kenapa 2 pangkat 0 sama dengan 1

Kenapa 2⁰ = 1 ?

2⁰ = 2^1-1
     = 2¹ x 2^-1
     = 2¹ : 2¹
     = 2 : 2
     = 1

2⁰ = 1

Logaritma

^alog b = p       artinya        a^p = b

dengan   0 < a < 1   atau   a > 1

untuk   a = 10,  a tidak ditulis, jadi log b maksudnya ¹⁰log b

untuk a = e (e = 2,7128....), ^elog a ditulis ln a (logaritma natural dari a)

 

 Sifat - sifat Logaritma

1. ^alog b + ^alog c = ^alog b c


2. ^alog b - ^alog c = ^alog $\frac{b}{c}$


3. ^alog bⁿ = n . ^alog b


4. ^alog b . ^blog c = ^alog c


5. ^alog b = $\frac{{}^{d}log\; b}{{}^{d}log\; a}$


6. ^alog b = ^anlog bⁿ


7. ^alog b = $\frac{1}{{}^{b}log\; a}$


8. ^amlog bⁿ = $\frac{m}{n}$ x ^alog b


9. a^{alog b} = b

 

Tambahan

1. log² a = log a x log a


2. log^-1 a = $\frac{1}{log\; a}$

Identitas Trigonometri

tan a = $\frac{sin\; a}{cos\; a}$

ctg a = $\frac{cos\; a}{sin\; a}$

sec a = $\frac{1}{cos\; a}$

cosec a = $\frac{1}{sin\; a}$

sin² a + cos² a = 1

1 + tan² a = sec² a

1 + ctg² a = cosec² a

Rumus Trigonometri Sudut Berelasi

Kuadran I

sin (90 - a) = cos a

cos (90 - a) = sin a

tan (90 - a) = ctg a

ctg (90 - a) = tan a 

Kuadran II

sin (180 - a) = sin a

cos (180 - a) = - cos a

tan (180 - a) = - tan a

ctg (180 - a) = - ctg a

sin (90 + a) = cos a

cos (90 + a) = - sin a

tan (90 + a) = - ctg a

ctg (90 + a) = - tan a

Kuadran III

sin (180 + a) = - sin a

cos (180 + a) = - cos a

tan (180 + a) = tan a

ctg (180 + a) = ctg a

sin (270 - a) = - cos a

cos (270 - a) = -sin a

tan (270 - a) = ctg a

ctg (270 - a) = tan a

Kuadran IV

sin (360 - a) = - sin a

cos (360 - a) = cos a

tan (360 - a) = - tan a

ctg (360 - a) = - ctg a

sin (270 + a) = cos a

cos (270 + a) = - sin a

tan (270 + a) = - ctg a

ctg (270 + a) = - tan a

Sudut Negatif 

sin (- a) = - sin a

cos (- a) = cos a

tan (- a) = - tan a

ctg (- a) = - ctg a

Pengulangan Sudut

sin (k . 360 + a) = sin a

cos (k . 360 + a) = cos a

tan (k . 360 + a) = tan a

ctg (k . 360 + a) = ctg a

Pangkat Rasional (Soal Jawab)

Soal

1. Sederhanakan bentuk bilangan berpangkat berikut :
   
   1. 2² x 2³
      
   2. (3²)⁵ . 3⁵
      
   3. ${\left(\frac{2}{3}\right)}^2.{\left(\frac{3}{2}\right)}^3$
      
   4. (0,125)² x (0,125)³
      
   5. (2⁴)² : 2³
      
   6. (5⁴)³ x 5⁵
      
   7. (3⁴ x 3³) x 3⁵
      
   8. (4⁶ + 8³) 2³
      
   9. (81³ - 243) + 16²
      
   10. (6² x 6³) x 6⁴
       
   11. ${\left(\frac{3}{4}\right)}^{4}x{\left(\frac{2}{3}\right)}^3$
       
2. Hitung hasilnya dengan mengubah menjadi bilangan berpangkat terlebih dahulu !
   1. $\sqrt[3]{\frac{1}{9}}x\sqrt[3]{\frac{1}{3}}x\frac{1}{3}\sqrt[4]{9}x9\sqrt[6]{8}$
   2. $(8^{-\frac{1}{3}}+8^{-2}-8^{-\frac{2}{3}}+8^{\frac{2}{3}}):\left(2^{-2}\right)$
   3. $\frac{{(16)}^{-\frac{3}{4}}+{(8)}^{-\frac{4}{3}}}{\mathrm{(2)}{(4)}^{-\frac{3}{2}}+8^{-1}}$
   4. $\frac{(a^{\frac{3}{2}}+a^{\frac{1}{2}})}{(a^{\frac{1}{2}}+a^{-\frac{1}{2}})}x\frac{(a^{\frac{1}{3}}+a^{-\frac{1}{3}})}{(\mathrm{a}+a^{\frac{1}{3}})}$

Jawab :

1. 1. 2² x 2³ = 2^{2 + 3} = 2⁵
      
   2. (3²)⁵ . 3⁵ = 3^{2 x 5} . 3⁵ = 3¹⁰ . 3⁵ = 3^{10 + 5} = 3¹⁵
      
   3. ${\left(\frac{2}{3}\right)}^2.{\left(\frac{3}{2}\right)}^3=\frac{2^2}{3^2}.\frac{3^3}{2^3}=\frac{2^2}{2^3}.\frac{3^3}{3^2}$ = 2^{2 - 3} . 3^{3 - 2} = 2^-1 . 3¹ = $\frac{3}{2}$
      
   4. (0,125)² x (0,125)³ = (0,125)^{2 + 3} = (0,125)⁵ = ${\left(\frac{1}{8}\right)}^5={\left(\frac{1}{2^3}\right)}^5$ = (2^-3)⁵ = 2^-15
   5. (2⁴)² : 2³ = 2^{4 x 2} : 2³ = 2⁸ : 2³ = 2^{8 - 3} = 2⁵
   6. (5⁴)³ x 5⁵ = 5^{4 x 3} x 5⁵ = 5¹² x 5⁵ = 5^{12 + 5} = 5¹⁷
   7. (3⁴ x 3³) x 3⁵ = (3^{4 + 3}) x 3⁵ = 3⁷ x 3⁵ = 3^{7 + 5} = 3¹²
   8. (4⁶ + 8³) 2³ = ((2²)⁶ + (2³)³) 2³ = (2¹² + 2⁹) 2³ = 2^{12 + 3} + 2^{9 + 3} = 2¹⁵ + 2¹²
   9. (81³ - 243) + 16² = ((3⁴)³ - 3⁵) + (2⁴)² = 3¹² - 3⁵ + 2⁸
   10. (6² x 6³) x 6⁴ = 6^{2 + 3} x 6⁴ = 6⁵ x 6⁴ = 6^{5 + 4} = 6⁹
   11. ${\left(\frac{3}{4}\right)}^{4}x{\left(\frac{2}{3}\right)}^3=\frac{3^4}{{\left(2^2\right)}^4}x\frac{2^3}{3^3}=\frac{3^4}{2^8}x\frac{2^3}{3^3}=\frac{3^4}{3^3}x\frac{2^3}{2^8}=3^{4\; -\; 3}x{2}^{3\; -\; 8}=3^{1}x{2}^{-5}=\frac{3}{2^5}$
       
2. 1. $\sqrt[3]{\frac{1}{9}}x\sqrt[3]{\frac{1}{3}}x\frac{1}{3}\sqrt[4]{9}x9\sqrt[6]{8}$
      
      = $3^{-\frac{2}{3}}x{3}^{-\frac{1}{3}}x{3}^{-1}{3}^{\frac{2}{4}}x{3}^2{2}^{\frac{3}{6}}$
      
      = $3^{-\frac{2}{3}-\frac{1}{3}-1+\frac{1}{2}}x{3}^2{2}^{\frac{1}{2}}$
      
      = $3^{-\frac{3}{2}+2}x{2}^{\frac{1}{2}}$
      
      = $3^{\frac{1}{2}}{2}^{\frac{1}{2}}$
      
      = $\sqrt{3}\sqrt{2}$
   2. $(8^{-\frac{1}{3}}+8^{-2}-8^{-\frac{2}{3}}+8^{\frac{2}{3}}):\left(2^{-2}\right)$
      
      = $({\left(2^3\right)}^{-\frac{1}{3}}+{\left(2^3\right)}^{-2}-{\left(2^3\right)}^{-\frac{2}{3}}+{\left(2^3\right)}^{\frac{2}{3}}):\left(2^{-2}\right)$
      
      = $(2^{-1}+2^{-6}-2^{-2}+2^2):\left(2^{-2}\right)$
      
      = $\left((,2^{-1},:,2^{-2},),+,(,2^{-6},:,2^{-2},),-,(,2^{-2},:,2^{-2},),+,(,2^2,:,2^{-2},)\right)$
      
      = $\left(2^{-1-(-2)},+,2^{-6-(-2)},-,2^{-2-(-2)},+,2^{2-(-2)}\right)$
      
      = $\left(2^1,+,2^{-4},-,2^0,+,2^4\right)$
      
      = $\left(2,+,2^{-4},-,1,+,2^4\right)$
      
      = $\left(1,+,2^{-4},+,2^4\right)$
      
      
   3. $\frac{{(16)}^{-\frac{3}{4}}+{(8)}^{-\frac{4}{3}}}{\mathrm{(2)}{(4)}^{-\frac{3}{2}}+8^{-1}}$
      
      = $\frac{{\left(2^4\right)}^{-\frac{3}{4}}+{\left(2^3\right)}^{-\frac{4}{3}}}{\mathrm{(2)}{\left(2^2\right)}^{-\frac{3}{2}}+{\left(2^3\right)}^{-1}}$
      
      = $\frac{2^{-3}+2^{-4}}{\mathrm{(2)}{2}^{-3}+2^{-3}}$
      
      = $\frac{\frac{1}{2^3}+\frac{1}{2^4}}{\frac{2}{2^3}+\frac{1}{2^3}}$
      
      = $\frac{\frac{2}{2^4}+\frac{1}{2^4}}{\frac{2}{2^3}+\frac{1}{2^3}}$
      
      = $\frac{\frac{3}{2^4}}{\frac{3}{2^3}}$
      
      = $\frac{3}{2^4}x\frac{2^3}{3}$
      
      = $\frac{1}{2}$
      
      
   4. $\frac{(a^{\frac{3}{2}}+a^{\frac{1}{2}})}{(a^{\frac{1}{2}}+a^{-\frac{1}{2}})}x\frac{(a^{\frac{1}{3}}+a^{-\frac{1}{3}})}{(\mathrm{a}+a^{\frac{1}{3}})}$
      
      = $\frac{(\mathrm{a\; .}{a}^{\frac{1}{2}}+a^{\frac{1}{2}})}{(a^{\frac{1}{2}}+\frac{1}{a^{\frac{1}{2}}})}x\frac{(a^{\frac{1}{3}}+\frac{1}{a^{\frac{1}{3}}})}{(\mathrm{a}+a^{\frac{1}{3}})}$
      
      = $\frac{\mathrm{a}(a^{\frac{1}{2}}+\frac{1}{a^{\frac{1}{2}}})}{(a^{\frac{1}{2}}+\frac{1}{a^{\frac{1}{2}}})}x\frac{a^{\frac{1}{3}}(1+\frac{1}{a^{\frac{2}{3}}})}{\mathrm{a}(1+\frac{1}{a^{\frac{2}{3}}})}$
      
      = $\mathrm{a}x\frac{a^{\frac{1}{3}}}{\mathrm{a}}$
      
      = $a^{\frac{1}{3}}$
      
      

Tabel Trigonometri Sudut - Sudut Istimewa

aº	sin aº	cos aº	tan aº
0º	0	1	0
30º	$\frac{1}{2}$	$\frac{1}{2}\sqrt{3}$	$\frac{1}{3}\sqrt{3}$
45º	$\frac{1}{2}\sqrt{2}$	$\frac{1}{2}\sqrt{2}$	1
60º	$\frac{1}{2}\sqrt{3}$	$\frac{1}{2}$	$\sqrt{3}$
90º	1	0	~

Contoh Soal :
Hitunglah nilai Sin 30º Cos 30º + Sin 60º Cos 60º

Jawab :
Sin 30º Cos 30º + Sin 60º Cos 60º = $\frac{1}{2}.\frac{1}{2}\sqrt{3}+\frac{1}{2}\sqrt{3}.\frac{1}{2}=\frac{1}{4}\sqrt{3}+\frac{1}{4}\sqrt{3}=\frac{1}{2}\sqrt{3}$


Soal Latihan 1

1. Hitunglah nilai dari :
   1. cos 60º + sin 30º + tan 45º
   2. tan 45º x cot 45º + sin 60º x cos 45º
   3. sin 60º x cos 30º - tan 45º x sin 45º
   4. cot 60º x cot 30º + tan 60º - cos 60º
   5. cos 30º x cos 30º + sin 30º x sin 30º


2. Diketahui segitiga ABC siku-siku di B. Jika panjang AC = 8 cm dan sudut A = 30º, hitunglah panjang AB dan BC !


3. Segitiga ABC siku-siku di C, titik D terletak pada AB sehingga CD tegak lurus AB, sudut ACD = 30º, dan AC = 10 cm. Hitunglah panjang AD, CD, BD dan BC !


4. Tunjukkan bahwa :
   1. sin² 30º + cos² 30º = 1
   2. $\frac{sin\; 60º}{cos\; 60º}=\mathrm{tan\; 60º}$
   3. $\frac{1}{cos\; 45º}=\mathrm{sec\; 45º}$
   4. 1 + tan² 60º = sec² 60º

Jawaban Soal Latihan UAS SMP Kelas VII

Soal dan Jawaban UAS SMP Kelas VII

1. Dari hasil survei yang dilakukan kepada 125 responden, diperoleh data sebagai berikut. 89 responden suka kopi, 96 responden suka teh, dan 68 responden suka kopi dan teh. Banyaknya responden yang tidak menyukai kopi dan teh adalah...
   Jawab :
   S = 125
   kopi dan teh = 68
   kopi = 89 - 68 = 21
   teh = 96 - 68 = 28
   
   
   
   
   tidak suka kopi dan teh = s - (kopi + kopi dan teh + teh) = 125 - (21 + 68 + 28) = 125 - 117 = 8
   
   
2. Perhatikan bentuk aljabar di bawah ini :
   -9ab² + 6bc² + 8ab²c + 6abc² + 2 - 6abc² + 9bc² - 5
    Selesaikan persamaan aljabar di atas.
   Jawab :
   -9ab² + 6bc² + 8ab²c + 6abc² + 2 - 6abc² + 9bc² - 5
   = -9ab² + 8ab²c + 6abc² - 6abc² + 6bc² + 9bc² + 2 - 5
   = -9ab² + 8ab²c + 15bc² - 3
   
   
3. Apabila a = -6 dan b = 8. Tentukan nilai dari -a² - 3a - 8b.
   Jawab :
   -a² - 3a - 8b = -(-6)² - 3(-6) - 8(8) = -(36) + 18 - 64 = -82
   
   
4. Jika a = 9x² + xy + 5y² dan b = 4x² - 7xy - y². Nilai dari b - a adalah.
   Jawab :
   b - a = 4x² - 7xy - y² - (9x² + xy + 5y²)
   = -5x² - 8xy - 6y²
   
   
5. Tentukan hasil perkalian (-3a - 2b)(a - 7b)
   Jawab :
   (-3a - 2b)(a - 7b) = -3a² + 21ab - 2ab + 14b² = -3a² + 19ab + 14b²
   
   
6. Hasil dari -9p(2p² + 9r) + 7(-4p -6pr) adalah
   Jawab :
   -9p(2p² + 9r) + 7(-4p -6pr) = -18p³ - 81pr - 28p - 42pr = -18p³ - 123pr - 28p
   
   
7. Diketahui r = (-5x² + 9y) dan s = (9x - 8y). Bentuk sederhana dari 7r - 2s adalah.
   Jawab :
   7r - 2s = 7(-5x² + 9y) - 2(9x - 8y) = -35x² + 63y - 18x + 16y = -35x² - 18x + 63y + 16y
   = -35x² - 18x + 79y = -x (35x + 18) + 79y
   
   
8. Diketahui
   $r=\frac{5x^2}{9z}$
   $t=\frac{7y^2}{12x^{2}yz}$
   Tentukan :
   
   1. penjumlahan r dan t
      
   2. perkalian r dan t
      
   3. hasil r dibagi t
      
   Jawab :
   
   1. r + t = $\frac{5x^2}{9z}+\frac{7y^2}{12x^{2}yz}=\frac{5x^2\left(12x^{2}yz\right)}{9z\left(12x^{2}yz\right)}+\frac{7y^2\left(9z\right)}{12x^{2}yz\left(9z\right)}$ = $\frac{60x^{4}yz}{108x^{2}y{z}^2}+\frac{63y^{2}z}{108x^{2}y{z}^2}$ = $\frac{60x^{4}yz+63y^{2}z}{108x^{2}y{z}^2}$
      
      
   2. r x t = $\frac{5x^2}{9z}\times \frac{7y^2}{12x^{2}yz}=\frac{35x^2{y}^2}{108x^{2}y{z}^2}$
      
      
   3. $\frac{r}{t}=\frac{\frac{5x^2}{9z}}{\frac{7y^2}{12x^{2}yz}}$ = $\frac{5x^2}{9z}\times \frac{12x^{2}yz}{7y^2}=\frac{60x^{4}yz}{63y^{2}z}$
      

Rumus Pangkat Rasional

RUMUS PANGKAT RASIONAL

1. a^m . aⁿ = a^{m + n}
   
2. $\frac{a^m}{a^n}=a^{m\; -\; n}$
   
3. (a^m . b^m) = (a . b)^m
   
4. $\frac{a^m}{b^m}={\left(\frac{a}{b}\right)}^m$
   
5. (a^m)ⁿ = a^mn
   
6. a⁰ = 1   , a ≠ 0
7. $a^{-m}=\frac{1}{a^m}$
   
8. $a^{\frac{m}{n}}=\sqrt[n]{a^m}$
   
9. $a^{-\frac{m}{n}}=\frac{1}{\sqrt[n]{a^m}}$
   
10. $\sqrt[m]{a}.\sqrt[m]{b}=\sqrt[m]{ab}={ab}^{\frac{1}{m}}$
11. $\frac{\sqrt[m]{a}}{\sqrt[m]{b}}=\sqrt[m]{\frac{a}{b}}={\left(\frac{a}{b}\right)}^{\frac{1}{m}}$

Contoh Soal :
Sederhanakan bentuk bilangan berpangkat berikut :

1. 2⁵ x 2²
   
2. $\frac{2^3}{2^2}$
   
3. (3²)³
   

Jawab :

1. 2⁵ x 2² = 2^{5 + 2} = 2⁷
   
2. $\frac{2^3}{2^2}=2^{3\; -\; 2}=2^1=2$
   
3. (3²)³ = 3⁶
   

Persamaan Kuadrat (Jawaban Soal Latihan 1) bagian 2

4. Selesaikan persamaan kuadrat berikut :
1. 5x² – 10x = 0
2. 2(x - 2)² = 8
3. 15x² + 16x + 4 = 0
4. $x^2-2x\sqrt{3}-1=0$
5. Selesaikan persamaan kuadrat berikut dengan cara yang paling tepat dan cepat :
1. (x - 1)(x - 2) = 12
2. (x + 1)(3x - 2) = 1 + x
3. $x+\frac{2}{x}=\frac{3+2x}{x}$
4. $\frac{x-6}{2}=-\frac{4}{x}$
5. $\sqrt{x+2}=x-4$
6. $\frac{x-\sqrt{3}}{x+\sqrt{3}}+\frac{x+\sqrt{3}}{x-\sqrt{3}}=3$
7. $\frac{x+1}{x+2}+\frac{x+2}{x+1}=2\frac{1}{2}$
8. $x+2+\sqrt{x+2}-6=0$
9. x⁴ - 13x² + 36 = 0
10. (x² + 1)² - 3(x² + 1) + 2 = 0

Jawab :

4. 1. 5x² – 10x = 0
      5x (x - 2) = 0
      5x = 0 atau x - 2 = 0
      x = 0 atau x = 2
      
   2. 2 (x - 2)² = 8
      (x - 2)² = $\frac{8}{2}$
      (x - 2)² = 4
      x - 2 = $\sqrt{4}$
      x - 2 = ± 2
      x = 2 ± 2
      x = 2 + 2 atau x = 2 - 2
      x = 4 atau x = 0
      
   3. 15x² + 16 x + 4 = 0
      (3x + 2)(5x + 2) = 0
      3x + 2 = 0 atau 5x + 2 = 0
      3x = -2 atau 5x = -2
      x = $-\frac{2}{3}$ atau x = $-\frac{2}{5}$
      
   4. $x^2-\mathrm{2x}\sqrt{3}-1=0$
      
      $x^2-2\sqrt{3}x-1=0$
      
      $x^2-2\sqrt{3}x=1$
      
      ${\left(x-\sqrt{3}\right)}^2=1+3$
      
      ${\left(x-\sqrt{3}\right)}^2=4$
      
      $x-\sqrt{3}=\sqrt{4}$
      
      $x-\sqrt{3}=\pm 2$
      
      $x=\sqrt{3}\pm 2$
      
      $x=\sqrt{3}+2$ atau $x=\sqrt{3}-2$
      
5. 1. (x-1)(x-2) = 12
      x² - 2x - x + 2 = 12
      x² - 3x + 2 -12 = 0
      x² - 3x - 10 = 0
      (x-5)(x+2) = 0
      x - 5 = 0 atau x + 2 = 0
      x = 5              x = -2
      
   2. (x+1)(3x-2) = 1 + x
      3x² - 2x + 3x - 2 = 1 + x
      3x² + x - 2 = 1 + x
      3x² + x - x - 2 - 1 = 0
      3x² - 3 = 0
      3x² = 3
      $x^2=\frac{3}{3}$
      x² = 1
      $x=\sqrt{1}$
      x = ± 1
      x = -1 atau x = 1
      
   3. $x+\frac{2}{x}=\frac{3+2x}{x}$
      x² + 2 = 3 + 2x
      x² - 2x + 2 - 3 = 0
      x² - 2x - 1 = 0
      x² - 2x = 1
      (x - 1)² = 1 + 1
      (x - 1)² = 2
      x - 1 = $\sqrt{2}$
      x - 1 = ±$\sqrt{2}$
      x = 1 ± $\sqrt{2}$
      x = 1 + $\sqrt{2}$ atau x = 1 - $\sqrt{2}$
      
   4. $\frac{x-6}{2}=-\frac{4}{x}$
      $\frac{x}{2}-\frac{6}{2}=-\frac{4}{x}$
      $\frac{1}{2}{x}^2-\mathrm{3x}=-4$
      $\frac{1}{2}{x}^2-\mathrm{3x}+4=0$
      x² - 6x + 8 = 0
      (x - 4)(x - 2) = 0
      x - 4 = 0 atau x - 2 = 0
      x = 4 atau x = 2
      
   5. $\sqrt{x+2}=x-4$
      x + 2 = (x - 4)²
      x + 2 = (x - 4)(x - 4)
      x + 2 = x² - 4x - 4x + 16
      x + 2 = x² - 8x + 16
      x² - 8x - x + 16 - 2 = 0
      x² - 9x + 14 = 0
      (x - 7)(x - 2) = 0
      x - 7 = 0 atau x - 2 = 0
      x = 7 atau x = 2
      
   6. $\frac{x-\sqrt{3}}{x+\sqrt{3}}+\frac{x+\sqrt{3}}{x-\sqrt{3}}=3$
      $\frac{x-\sqrt{3}}{x+\sqrt{3}}\frac{x-\sqrt{3}}{x-\sqrt{3}}+\frac{x+\sqrt{3}}{x-\sqrt{3}}\frac{x+\sqrt{3}}{x+\sqrt{3}}=3$
      $\frac{x^2-\sqrt{3}x-\sqrt{3}x+3}{x^2-\sqrt{3}x+\sqrt{3}x-3}+\frac{x^2+\sqrt{3}x+\sqrt{3}x+3}{x^2-\sqrt{3}x+\sqrt{3}x-3}=3$
      $\frac{x^2-2\sqrt{3}x+3}{x^2-3}+\frac{x^2+2\sqrt{3}x+3}{x^2-3}=3$
      $\frac{x^2+x^2-2\sqrt{3}x+2\sqrt{3}x+3+3}{x^2-3}=3$
      $\frac{2x^2+6}{x^2-3}=3$
      2x² + 6 = 3 ( x² - 3 )
      2x² + 6 = 3x² - 9
      3x² - 2x² - 9 - 6 = 0
      x² - 15 = 0
      x² = 15
      x = $\sqrt{15}$
      x = $\pm \sqrt{15}$
      x = $\sqrt{15}$ atau x = $-\sqrt{15}$
      
   7. $\frac{x+1}{x+2}+\frac{x+2}{x+1}=2\frac{1}{2}$
      $\frac{\mathrm{(x\; +\; 1)}\mathrm{(x\; +\; 1)}}{\mathrm{(x\; +\; 2)}\mathrm{(x\; +\; 1)}}+\frac{\mathrm{(x\; +\; 2)}\mathrm{(x\; +\; 2)}}{\mathrm{(x\; +\; 1)}\mathrm{(x\; +\; 2)}}=\frac{5}{2}$
      $\frac{x^2+\mathrm{2x}+1}{x^2+\mathrm{3x}+2}+\frac{x^2+\mathrm{4x}+4}{x^2+\mathrm{3x}+2}=\frac{5}{2}$
      $\frac{x^2+x^2+\mathrm{2x}+\mathrm{4x}+1+4}{x^2+\mathrm{3x}+2}=\frac{5}{2}$
      $\frac{2x^2+\mathrm{6x}+5}{x^2+\mathrm{3x}+2}=\frac{5}{2}$
      $2x^2+\mathrm{6x}+5=\frac{5}{2}{x}^2+\mathrm{3x}+2$
      $4x^2+\mathrm{12x}+10=5x^2+\mathrm{15x}+10$
      $5x^2-4x^2+\mathrm{15x}-\mathrm{12x}+10-10=0$
      x² + 3x = 0
      x(x + 3) = 0
      x = 0 atau x + 3 = 0
                      x = 3
      
   8. $x+2+\sqrt{x+2}-6=0$
      $x+2-6+\sqrt{x+2}=0$
      $x-4+\sqrt{x+2}=0$
      $\sqrt{x+2}=\mathrm{-x}+4$
      x + 2 = (-x + 4)²
      x + 2 = (-x + 4)(-x + 4)
      x + 2 = x² - 4x - 4x + 16
      x + 2 = x² - 8x + 16
      x² - 8x - x + 16 - 2 = 0
      x² - 9x + 14 = 0
      (x - 7)(x - 2) = 0
      x - 7 = 0 atau x - 2 = 0
      x = 7 atau x = 2
      
   9. x⁴ - 13x² + 36 = 0
      (x² - 4)(x² - 9) = 0
      x² - 4 = 0 atau x² - 9 = 0
      x² = 4 atau x² = 9
      x = $\sqrt{4}$ atau x = $\sqrt{9}$
      x = ±2 atau x = ±3
      x = 2 atau x = -2 atau x = 3 atau x = -3
      
   10. (x² + 1)² - 3(x² + 1) + 2 = 0
       misal : a = x² + 1
       a² - 3a + 2 = 0
       (a - 2)(a - 1) = 0
       a - 2 = 0 atau a - 1 = 0
       a = 2 atau a = 1
       x² + 1 = 2 atau x² + 1 = 1
       x² = 2 - 1 atau x² = 1 - 1
       x² = 1 atau x² = 0
       x = $\sqrt{1}$ atau x = $\sqrt{0}$
       x = ±1 atau x = 0
       x = 1 atau x = -1 atau x = 0
       

Persamaan Kuadrat (Jawaban Soal Latihan 1)

Soal Latihan 1

1. Selesaikan persamaan kuadrat berikut dengan cara pemfaktoran :
1. x² – 4 = 0
2. x² – 5 = 0
3. x² – 3x = 0
4. 3x - 4x² = 0
5. x² + 4x - 12 = 0
6. 2x² + 5x - 12 = 0
7. 6x² – x - 15 = 0
8. -x² – 5x + 24 = 0
9. $2x^2-\genfrac{}{}{0.1ex}{}{1}{3}x-5=0$
10. $25+\frac{7}{x}=12x$
2. Selesaikan persamaan kuadrat berikut dengan cara melengkapkan kuadrat sempurna :
1. x² – x – 12 = 0
2. x² – 2x – 8 = 0
3. 2x² – 6x + 3 = 0
4. 3x² = 4x + 6
3. Selesaikan persamaan kuadrat berikut dengan menggunakan rumus abc :
1. x² – 5x – 9 = 0
2. 2x² + 5x – 12 = 0
3. 3x² – 8x - 3 = 0
4. 6 - 3x - 2x² = 0
5. 4x² – 5ax + a² = 0
4. Selesaikan persamaan kuadrat berikut :
1. 5x² – 10x = 0
2. 2(x - 2)² = 8
3. 15x² + 16x + 4 = 0
4. $x^2-2x\sqrt{3}-1=0$
5. Selesaikan persamaan kuadrat berikut dengan cara yang paling tepat dan cepat :
1. (x - 1)(x - 2) = 12
2. (x + 1)(3x - 2) = 1 + x
3. $x+\frac{2}{x}=\frac{3+2x}{x}$
4. $\frac{x-6}{2}=-\frac{4}{x}$
5. $\sqrt{x+2}=x-4$
6. $\frac{x-\sqrt{3}}{x+\sqrt{3}}+\frac{x+\sqrt{3}}{x-\sqrt{3}}=3$
7. $\frac{x+1}{x+2}+\frac{x+2}{x+1}=2\frac{1}{2}$
8. $x+2+\sqrt{x+2}-6=0$
9. x⁴ - 13x² + 36 = 0
10. (x² + 1)² - 3(x² + 1) + 2 = 0

Jawab :

1. 1. x² - 4 = 0        #cara pemfaktoran : cari angka dikali -4 dijumlah 0 jadi angkanya 2 dan -2
      (x + 2)(x - 2) = 0
      x + 2 = 0   atau   x - 2 = 0
      x = -2         atau x = 2
   2. x² – 5 = 0       #dikali –5 dijumlah 0 jadi angkanya √5 dan -√5
      (x - √5)(x + √5) = 0
      x - √5 = 0  atau   x + √5 = 0
      x = √5                 x = -√5
      
   3. x2 – 3x = 0
      x(x – 3) = 0
      x = 0 atau x – 3 = 0
                       x = 3
   4. 3x – 4x² = 0
      x(3 - 4x) = 0
      x = 0 atau 3 - 4x = 0
                       3 = 4x
                       x = $\frac{3}{4}$
      
   5. x² + 4x – 12 = 0
      (x + 6 )(x – 2) = 0
      x = -6 atau x = 2
      
   6. 2x² + 5x – 12 = 0
      (2x - 3)(x + 4) = 0
      2x – 3 = 0  atau  x + 4 = 0
      2x = 3                 x = -4
      x = $\frac{3}{2}$
   7. 6x² – x – 15 = 0
      (2x + 3)(3x - 5) = 0
      2x + 3 = 0  atau  3x – 5 = 0
      2x = -3                3x = 5
      x = $-\frac{3}{2} \; \; \; \; \; \; \; \; \;$ x = $\frac{5}{3}$
   8. -x² - 5x + 24 = 0
      x² + 5x – 24 = 0
      (x + 8)(x – 3) = 0
      x + 8 = 0  atau  x – 3 = 0
      x = -8                x = 3
   9. 2x² - $\frac{1}{3}x$ – 5 = 0
      6x² – x – 15 = 0
      (2x + 3)(3x - 5) = 0
      2x + 3 = 0  atau  3x – 5 = 0
      2x = -3                3x = 5
      x = $-\frac{3}{2}$                 x = $\frac{5}{3}$
   10. 25 + $\frac{7}{x}$ = 12x
       25x + 7 = 12x²
       12x² - 25x – 7 = 0
       (3x – 7)(4x + 1) = 0
       3x – 7 = 0  atau  4x + 1 = 0
       3x = 7                 4x = -1
       x = $\frac{7}{3}$                   x = $-\frac{1}{4}$
2. Melengkapkan kuadrat sempurna :
   
   
   1. x² – x – 12 = 0
      x² – x = 12                #nilai b dibagi 2
      (x - $\frac{1}{2}$)² = 12 + $\frac{1}{4}$
      (x - $\frac{1}{2}$)² = $\frac{49}{4}$
      $x-\frac{1}{2}=\sqrt{\frac{49}{4}}$
      $x-\frac{1}{2}=\pm \frac{7}{2}$
      $x=\frac{1}{2}\pm \frac{7}{2}$
      $x=\frac{1}{2}+\frac{7}{2}$  atau  $x=\frac{1}{2}-\frac{7}{2}$
      x = $\frac{8}{2}$               x = $-\frac{6}{2}$
      x = 4                 x = -3
   2. x² – 2x – 8 = 0
      x² – 2x = 8
      (x – 1)² = 8 + 1
      (x – 1)² = 9
      x – 1 = √9
      x – 1 = ± 3
      x = 1 ± 3
      x = 1 + 3 atau x = 1 – 3
      x = 4               x = -2
   3. 2x² – 6x + 3 = 0
      2x² – 6x = -3
      x² – 3x = $-\frac{3}{2}$
      ${\left(x-\frac{3}{2}\right)}^2=-\frac{3}{2}+\frac{9}{4}$
      ${\left(x-\frac{3}{2}\right)}^2=-\frac{6}{4}+\frac{9}{4}$
      ${\left(x-\frac{3}{2}\right)}^2=\frac{3}{4}$
      $x-\frac{3}{2}=\sqrt{\frac{3}{4}}$
      $x-\frac{3}{2}=\pm \frac{\sqrt{3}}{\sqrt{4}}$
      $x-\frac{3}{2}=\pm \frac{\sqrt{3}}{2}$
      $x-\frac{3}{2}=\pm \frac{1}{2}\sqrt{3}$
      $x=\frac{3}{2}\pm \frac{1}{2}\sqrt{3}$
      $x=\frac{3}{2}+\frac{1}{2}\sqrt{3}$  atau  $x=\frac{3}{2}-\frac{1}{2}\sqrt{3}$
   4. 3x² = 4x + 6
      3x² - 4x = 6
      x² - $\frac{4}{3}x$ = 2
      ${\left(x-\frac{2}{3}\right)}^2=2+\frac{4}{9}$
      ${\left(x-\frac{2}{3}\right)}^2=\frac{22}{9}$
      $x-\frac{2}{3}=\sqrt{\frac{22}{9}}$
      $x-\frac{2}{3}=\pm \frac{1}{3}\sqrt{22}$
      $x=\frac{2}{3}\pm \frac{1}{3}\sqrt{22}$
      $x=\frac{2}{3}+\frac{1}{3}\sqrt{22}$   atau   $x=\frac{2}{3}-\frac{1}{3}\sqrt{22}$
3. Menggunakan Rumus abc :
   
   
   1. x² – 5x – 9 = 0
      Rumus abc : $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
      a = 1, b = -5, c = -9
      $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
      $x=\frac{-\mathrm{(-5)}\pm \sqrt{{\mathrm{(-5)}}^2-4\mathrm{(1)}\mathrm{(-9)}}}{2\mathrm{(1)}}$
      $x=\frac{5\pm \sqrt{25+36}}{2}$
      $x=\frac{5\pm \sqrt{61}}{2}$
      $x=\frac{5+\sqrt{61}}{2}$ atau $x=\frac{5-\sqrt{61}}{2}$
   2. 2x² + 5x – 12 = 0
      a = 2, b = 5, c = -12
      $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
      $x=\frac{-5\pm \sqrt{{\mathrm{(5)}}^2-4\mathrm{(2)}\mathrm{(-12)}}}{2\mathrm{(2)}}$
      $x=\frac{-5\pm \sqrt{25+96}}{4}$
      $x=\frac{-5\pm \sqrt{121}}{4}$
      $x=\frac{-5\pm 11}{4}$
      $x=\frac{-5+11}{4}$ atau $x=\frac{-5-11}{4}$
      $x=\frac{6}{4}$        atau $x=\frac{-16}{4}$
      $x=\frac{3}{2}$        atau x = -4
   3. 3x² - 8x – 3 = 0
      
      a = 3, b = -8, c = -3
      
      $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
      
      $x=\frac{-\mathrm{(-8)}\pm \sqrt{{\mathrm{(-8)}}^2-4\mathrm{(3)}\mathrm{(-3)}}}{2\mathrm{(3)}}$
      
      $x=\frac{8\pm \sqrt{64+36}}{6}$
      
      $x=\frac{8\pm \sqrt{100}}{6}$
      
      $x=\frac{8\pm 10}{6}$
      
      $x=\frac{8+10}{6}$ atau $x=\frac{8-10}{6}$
      
      $x=\frac{18}{6}$ atau $x=\frac{-2}{6}$
      
      x = 3 atau $x=-\frac{1}{3}$
   4. 6 - 3x - 2x² = 0
      
      -2x² - 3x + 6 = 0
      
      a = -2, b = -3, c = 6
      
      $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
      
      $x=\frac{-\mathrm{(-3)}\pm \sqrt{{\mathrm{(-3)}}^2-4\mathrm{(-2)}\mathrm{(6)}}}{2\mathrm{(-2)}}$
      
      $x=\frac{3\pm \sqrt{9+48}}{-4}$
      
      $x=\frac{3\pm \sqrt{57}}{-4}$
      
      $x=\frac{3+\sqrt{57}}{-4}$ atau $x=\frac{3-\sqrt{57}}{-4}$
      
      $x=-\frac{3}{4}-\frac{1}{4}\sqrt{57}$ atau $x=-\frac{3}{4}+\frac{1}{4}\sqrt{57}$
   5. 4x² - 5ax + a² = 0
      
      a = 4, b = -5a, c = a²
      
      $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
      
      
      $x=\frac{-\mathrm{(-5a)}\pm \sqrt{{\mathrm{(-5a)}}^2-4\mathrm{(4)}\left({\mathrm{a}}^2\right)}}{2\mathrm{(4)}}$
      
      
      $x=\frac{\mathrm{5a}\pm \sqrt{25{\mathrm{a}}^2-16{\mathrm{a}}^2}}{8}$
      
      
      $x=\frac{\mathrm{5a}\pm \sqrt{9{\mathrm{a}}^2}}{8}$
      
      
      $x=\frac{\mathrm{5a}\pm \mathrm{3a}}{8}$
      
      $x=\frac{\mathrm{5a}+\mathrm{3a}}{8}$ atau $x=\frac{\mathrm{5a}-\mathrm{3a}}{8}$
      
      $x=\frac{\mathrm{8a}}{8}$ atau $x=\frac{\mathrm{2a}}{8}$
      
      x = a atau $x=\frac{1}{4}\mathrm{a}$

Jawaban Soal nomer berikutnya dilanjutkan ke halaman selanjutnya klik disini