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4x^2+4x-3
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\(4x^2+4x-3\)
\(=4x^2+6x-2x-3\)
\(=\left(4x^2+6x\right)-\left(2x+3\right)\)
\(=2x\left(2x+3\right)-\left(2x+3\right)\)
\(=\left(2x+3\right)\left(2x-1\right)\)
Ta có: \(4x^2+12x+9\)
\(=4x^2+6x+6x+9\)
\(=2x\left(2x+3\right)+3\left(2x+3\right)\)
\(=\left(2x+3\right)^2\)
\(=4\left(x-2y\right)-\left(x-2y\right)\left(x+2y\right)\\ =\left(x-2y\right)\left(4-x-2y\right)\)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(4x^2-4x+1=\left(2x\right)^2-2.2x.1+1^2=\left(2x-1\right)^2\\ ---\\ 4x^2-4x-3\\ =4x^2-4x+1-4\\ =\left(2x-1\right)^2-2^2=\left(2x-1-2\right)\left(2x-1+2\right)\\ =\left(2x-3\right)\left(2x+1\right)\)
1: =(2x)^2-2*2x*1+1^2
=(2x-1)^2
2: =4x^2-6x+2x-3
=2x(2x-3)+(2x-3)
=(2x-3)(2x+1)
\(4x^2+4x-3\)
\(4x^2+4x+1-4\)
\(\left(2x+1\right)^2-2^2\)
\(\left(2x+1-2\right)\left(2x+1+2\right)\)
\(\left(2x-1\right)\left(2x+3\right)\)
\(4x^2+4x-3=\left(2x-1\right)\left(2x+3\right)\)
Ta có \(4x^2+4x-3=4x^2-2x+6x-3\)
\(=2x\left(2x-1\right)+3\left(2x-1\right)\)
\(=\left(2x-1\right)\left(2x+3\right)\)
Vậy \(4x^2+4x-3=\left(2x-1\right)\left(2x+3\right)\)