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x2+4x+3=x2+x+3x+3=(x2+x)+(3x+3)=x(x+1)+3(x+1)=(x+3)(x+1)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
4x2+4x-3
= 4x2+4x+1-4
=(2x+1)2-22
=(2x+1-2)(2x+1+2)
=(2x-1)(2x+3)
\(4x^2+4x-3\)
\(= 4x^2 + 4x + 1 -4\)
\(= ( 2x+1)^2 -2^2\)
\(= (2x+1-2)(2x+1+2)\)
\(= (2x-1)(2x+3)\)
\(x^3+4x^2+4x+1\)
\(=x^3+3x^2+x+x^2+3x+1\)
\(=x\left(x^2+3x+1\right)+\left(x^2+3x+1\right)\)
\(=\left(x+1\right)\left(x^2+3x+1\right)\)
\(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=\left(x+1\right)\left(4x^3-x\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left[\left(2x\right)^2-1\right]\)
\(=x\left(x+1\right)\left(2x+1\right)\left(2x-1\right)\)
(Nhớ k cho mình với nhá!)
\(4x^2+4x-3\)
\(4x^2+4x+1-4\)
\(\left(2x+1\right)^2-2^2\)
\(\left(2x+1-2\right)\left(2x+1+2\right)\)
\(\left(2x-1\right)\left(2x+3\right)\)
Trả lời:
4x2 + 4x - 3
= [ ( 2x )2 + 2.2x.1 + 1 ] - 4
= ( 2x + 1 )2 - 4
= ( 2x + 1 - 2 ) ( 2x + 1 + 2 )
= ( 2x - 1 ) ( 2x + 3 )