a, x^2-4=8(x-2)

=> x^2 - 4 = 8.x - 16

=> x^2 = (8.x - 16) - 4 

=> x^2 = 8.x - (16+4)

=> x^2 = 8.x - 20

A, \(x^2-4=8\left(x-2\right)\)=> \(\left(x-2\right).\left(x+2\right)=8\left(x-2\right)=>\left(x-2\right).\left(x+2\right)-8\left(x-2\right)=0\)

=>\(\left(x-2\right).\left(x-6\right)=0\)

=> x = 2 hoặc x =6 

B. \(x^2-4x+4=9\left(x-2\right)\)=> \(\left(x-2\right)^2=9\left(x-2\right)=>\left(x-2\right)^2-9\left(x-2\right)=0\)

=>\(\left(x-2\right).\left(x-11\right)=0\)=> x =2 hoặc x =11

C. \(4x^2-12x+9=\left(5-x\right)^2=>\left(2x-3\right)^2=\left(5-x\right)^2\)

=>\(\left(2x-3\right)^2-\left(5-x\right)^2=>\left(3x-8\right).\left(x+2\right)=0\)

=> x = 3/8 hoặc x = - 2

b: 

1: \(\Leftrightarrow2x\left(x+2\right)=0\)

=>x=0 hoặc x=-2

\(4x\left(12x-9\right)-8x\left(6x-5\right)=1\)

\(\Leftrightarrow48x^2-36x-48x^2+40x=1\)

\(\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\)

( 4x - 1 )³ + ( 3 - 4x )( 9 + 12x + 16x² ) = ( 8x - 1 )( 8x + 1 ) - ( 3x - 5 )

<=> 64x³ - 48x² + 12x - 1 + [ 3³ - ( 4x )³ ] = ( 8x )² - 1 - 3x + 5

<=> 64x³ - 48x² + 12x - 1 + 27 - 64x³ = 64x² - 3x + 4

<=> -48x² + 12x + 26 = 64x² - 3x + 4

<=> -48x² + 12x + 26 - 64x² + 3x - 4 = 0

<=> -112x² + 15x + 22 = 0 (*)

\(\Delta=b^2-4ac=15^2-4\cdot\left(-112\right)\cdot22=225+9856=10081\)

\(\Delta>0\)nên (*) có hai nghiệm phân biệt

\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{\sqrt{10081}-15}{-224}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-15-\sqrt{10081}}{-224}\end{cases}}\)

Lớp 8 sao nghiệm xấu thế -..-

Dạ em cảm ơn

a)x² +5x =0

=>x.x +5x =0

=> x.(5+x) =0

=>Hoặc 5+x =0 =>x= -5

Hoặc x= 0

Vậy x=-5 ; x=0

a) (x - 4)² - 36 = 0

=> (x - 4)² = 36

=> x - 4 = 6 hoặc x - 4 = -6

=> x = 10 hoặc x = -2

b) hình như sai đề bn ạ

c) x(x - 5) - 4x + 20 = 0

=> x(x - 5) - 4(x - 5) = 0

=> (x - 5)(x - 4) = 0

=> x - 5 = 0 hoặc x - 4 = 0

=> x = 5 hoặc x = 4