Trả lời:

sửa: \(x^3-4x^2-8x+8\)

\(=\left(x^3+8\right)-\left(4x^2+8x\right)\)

\(=\left(x+2\right)\left(x^2-2x+4\right)-4x\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-2x+4-4x\right)\)

\(=\left(x+2\right)\left(x^2-6x+4\right)\)

Bài 2 : 

 \(x^2-x-6=0\Leftrightarrow x^2-3x+2x-6=0\)

\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow x=-2;x=3\)

Ta có : \(A=x^4+2x^3+2x^2+2x+1=\left(x^2+1\right)^2+2x\left(x^2+1\right)\)

\(=\left(x^2+1\right)\left(x^2+2x+1\right)=\left(x^2+1\right)\left(x+1\right)^2\)

Với x = -2 thì A = \(\left(4+1\right)\left(-2+1\right)^2=5\)

Bài 4 : 

\(a^2+b^2=2ab\Leftrightarrow a^2-2ab+b^2=0\Leftrightarrow\left(a-b\right)^2=0\)

Đẳng thức xảy ra khi a = b 

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(x-1)+(x+3)=x-1+x+3=2x+2=2(x+1)

ko có kết quả bằng bao nhiêu thì sao tìm x đc

\(\frac{3\left(x-2\right)}{4}\div\frac{2-x}{2}=\frac{3\left(x-2\right)}{4}\times\frac{-2}{x-2}=\frac{-3}{2}\)

học tốt

Rút gọn nhé !

\(\frac{3}{4}.\left(x-2\right):\frac{1}{2}.\left(2-x\right)=\frac{3x-6}{4}.2.\left(2-x\right)\)

\(=\frac{3x-6}{4}.\left(4-2x\right)=\frac{\left(3x-6\right).\left(4-2x\right)}{4}\)

\(=\frac{\left(12x-24\right)-\left(6x^2+12x\right)}{4}=\frac{-24-6x^2}{4}\)

\(=\frac{-12-3x^2}{2}=\frac{-3.\left(4+x^2\right)}{2}\)