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= 4x2-y2+8y-16
= 4x2- (y2-8y+16)
= 4x2- (y-4)2
=(4x-y+4) (4x+y-4)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(4x^2+4x-3\)
\(4x^2+4x+1-4\)
\(\left(2x+1\right)^2-2^2\)
\(\left(2x+1-2\right)\left(2x+1+2\right)\)
\(\left(2x-1\right)\left(2x+3\right)\)
\(=x\left(1-4x^2\right)=x\left(1-2x\right)\left(1+2x\right)\)
4x2+4x-3
= 4x2+4x+1-4
=(2x+1)2-22
=(2x+1-2)(2x+1+2)
=(2x-1)(2x+3)
\(4x^2+4x-3\)
\(= 4x^2 + 4x + 1 -4\)
\(= ( 2x+1)^2 -2^2\)
\(= (2x+1-2)(2x+1+2)\)
\(= (2x-1)(2x+3)\)
\(x^3+4x^2+4x+1\)
\(=x^3+3x^2+x+x^2+3x+1\)
\(=x\left(x^2+3x+1\right)+\left(x^2+3x+1\right)\)
\(=\left(x+1\right)\left(x^2+3x+1\right)\)
4x2+4x-3
= 4x2-2x+6x-3
= 2x*(2x-1)+3*(2x-1)
= (2x+3)*(2x-1)
\(4x^2+4x-3=\left(2x\right)^2+2.\left(2x\right).1+1^2-4=\left(2x+1\right)^2-2^2=\left(2x+1-2\right)\left(2x+1+2\right)=\left(2x-1\right)\left(2x+3\right)\)