Phân tích thành nhân tử:
x 2 - 6
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\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
(3x+2)2-(x-6)2=(3x+2-x+6)(3x+2+x-6)=(2x+8)(4x-4)=8(x+4)(x-1)
\(=x\left(x-6\right)+2\left(x-6\right)=\left(x-6\right)\left(x+2\right)\)
\(\left(3x+2\right)^2-\left(x-6\right)^2=\left(3x+2-x+6\right)\left(3x+2+x-6\right)=\left(2x+8\right)\left(4x-4\right)=8\left(x+4\right)\left(x-1\right)\)
\(\left(3x+2\right)^2-\left(x-6\right)^2\)
\(=\left(3x+2+x-6\right)\left(3x+2-x+6\right)\)
\(=\left(4x-4\right)\left(2x+8\right)\)
\(=8\left(x-1\right)\left(x+4\right)\)
\(\left(x^2+1\right)^2-6\left(x^2+1\right)+9\)
\(=\left[\left(x^2+1\right)-3\right]^2\)
\(=\left(x^2+1-3\right)^2\)
\(=\left(x^2-2\right)^2\)
\(x^2+x-2\)
\(=x^2-x+2x-2\)
\(=x\left(x-1\right)+2\left(x-1\right)\)
\(=\left(x-1\right)\left(x+2\right)\)
chì làm được 1 cách thôi
\(x^2-6x+8\)
\(=\left(x^2-6x+9\right)-1\)
\(=\left(x-3\right)^2-1^2\)
\(=\left(x-3-1\right)\left(x-3+1\right)\)
\(=\left(x-4\right)\left(x-2\right)\)
(Tíck cho mìk vs nha!)
cách 2:
x2 -6x +8 = x2 -2x -4x+8= x(x-2) -4(x-2)
= (x-2)(x-4)
x2 - 6 = x2 - (√6)2 = (x - √6)(x + √6)