\(x^4-8x^3+11x^2+8x-12=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(x^2-8x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-6\right)\left(x-2\right)=0\)

\(\Leftrightarrow x=\left\{1;-1;6;2\right\}\)

\(x^4-8x^3+11x^2+8x-12=0\)

\(\Leftrightarrow x^4-x^3-7x^3+7x^2+4x^2-4x+12x-12=0\)

\(\Leftrightarrow\left(x^4-x^3\right)-\left(7x^3-7x^2\right)+\left(4x^2-4x\right)+\left(12x-12\right)=0\)

\(\Leftrightarrow x^3\left(x-1\right)-7x^2\left(x-1\right)+4x\left(x-1\right)+12\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-7x^2+4x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2-8x^2-8x+12x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)[x^2\left(x+1\right)-8x\left(x+1\right)+12\left(x+1\right)]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^2-8x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x-6\right)=0\)

\(\Leftrightarrow\)x - 1 =0 ; x + 1 = 0 ; x - 2 =0 hoặc x - 6 = 0

\(\Leftrightarrow\)x = 1 ; x = -1 ; x = 2 ; x=6

x.(8x - 2) - 8x² + 12 = 0

=> 8x² - 2x - 8x² + 12 = 0

=> 12 - 2x = 0

=> 2x = 12

=> x = 12 : 2

=> x = 6

Vậy x = 6

a,\(x\left(8x-2\right)-8x^2+12=0\)

\(\Leftrightarrow8x^2-2x-8x^2+12=0\)

\(\Leftrightarrow-2x+12=0\)

\(\Leftrightarrow-2x=-12\)

\(\Leftrightarrow x=6\)

b,\(x\left(4x-4\right)-\left(2x+1\right)^2=0\)

\(\Leftrightarrow4x^2-5x-\left(4x^2+4x+1\right)=0\)

\(\Leftrightarrow4x^2-5x-4x^2-4x-1=0\)

\(\Leftrightarrow-9x-1=0\)

\(\Leftrightarrow-9x=1\)

\(\Leftrightarrow x=\frac{-1}{9}\)

A:x(8x -2) -8x²+12=0

8x²-2x-8x²+12=0

-2x+12=0

-2x=-12

x=6

Vậy......

b:x(4x-5)-(2x+1)²=0

4x²-5x-4x²-4x-1=0

-9x=1

x=-1/9

Vậy....

a)   \(x^2+8x+16=0\)

<=>  \(\left(x+4\right)^2=0\)

<=> \(x=-4\)

Vậy...

b) mk chỉnh đề

 \(4x^2-12x=-9\)

<=>  \(4x^2-12x+9=0\)

<=>  \(\left(2x-3\right)^2=0\)

<=>  \(x=\frac{3}{2}\)

Vậy....

a﴿ x + 8x + 16 = 0

<=> x + 4 = 0

<=> x = −4

Vậy x=-4

\(P=1+\frac{x+3}{x^2+5x+6}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)

\(P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)

\(P=1+\frac{1}{x+2}:\left(\frac{4x^2.2}{4x^2\left(x-2\right)}-\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{1}{x+2}\right)\)

\(P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right)\)

\(P=1+\frac{1}{x+2}:\left(\frac{2x+4-x-x+2}{\left(x+2\right)\left(x-2\right)}\right)\)

\(P=1+\frac{1}{x+2}:\frac{6}{\left(x+2\right)\left(x-2\right)}=1+\frac{\left(x+2\right)\left(x-2\right)}{6\left(x+2\right)}=1+\frac{x-2}{6}\)

\(=\frac{x+4}{6}.P=0\Leftrightarrow x=-4\)

\(P>0\Leftrightarrow x>-4\)

sai lớp :>>>

\(a,x\left(8x-2\right)-8x^2+12=0\)

\(\Rightarrow8x^2-2x-8x^2+12=0\)

\(\Rightarrow-2x+12=0\)

\(\Rightarrow-2x=-12\)

\(\Rightarrow x=6\)

\(b,x\left(4x-5\right)-\left(2x+1\right)^2=0\)

\(\Rightarrow4x^2-5x-4x^2-4x-1=0\)

\(\Rightarrow-9x-1=0\)

\(\Rightarrow-9x=1\)

\(\Rightarrow x=\frac{-1}{9}\)

a) x(8 - 2) - 8x² + 12 = 0

x(8 - 2) - 8x² = 12 - 0

x(8 - 2) - 8x² = 12

2x = 12

x = 6

b) x(4x - 5) - (2x + 1)² = 0

9x - 1 = 0

9x = 0 + 1

9x = 1

x = -1/9

\(a,x\left(8x-2\right)-8x^2+12=0\)

\(\Leftrightarrow8x^2-2x-8x^2+12=0\)

\(\Leftrightarrow-2x+12=0\)

\(\Leftrightarrow-2x=-12\)

\(\Leftrightarrow x=6\)

\(b,x\left(4x-5\right)-\left(2x+1\right)^2=0\)

\(\Leftrightarrow4x^2-5x-4x^2-4x-1=0\)

\(\Leftrightarrow-9x-1=0\)

\(\Leftrightarrow-9x=1\)

\(\Leftrightarrow x=-\frac{1}{9}\)

(x-2)(x-6)=0

x-2=0 hoặc x-6=0

x=2 hoắc x=6

\(\Rightarrow x=2\)

Chỉ pt tới dok thuj!^^

\(x^2-8x+12=0\)

=>\(x^2-2x-6x+12=0\)

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

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

=>\(\orbr{\begin{cases}x-6=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=6\\x=2\end{cases}}}\)

Vậy ...