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\(4x^4+4x^3+5x^2+2x+1\)
\(=4x^4+2x^3+2x^2+2x^3+x^2+2x^2+x+1\)
\(=2x^2\left(2x^2+x+1\right)+x\left(2x^2+x+1\right)+\left(2x^2+x+1\right)\)
\(=\left(2x^2+x+1\right)\left(2x^2+x+1\right)\)
\(=\left(2x^2+x+1\right)^2\)
C1: \(4x^2-4x+1=\left(2x-1\right)^2\) (Hằng đẳng thức bạn ạ)
C2: \(4x^2-4x+1\)
=\(4x^2-2x-2x+1\)
=\(2x\left(2x-1\right)-\left(2x-1\right)\)
=\(\left(2x-1\right)\left(2x-1\right)\)
=\(\left(2x-1\right)^2\)
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\(x^3+4x^2+4x+1\)
\(=x^3+3x^2+x+x^2+3x+1\)
\(=x\left(x^2+3x+1\right)+\left(x^2+3x+1\right)\)
\(=\left(x+1\right)\left(x^2+3x+1\right)\)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(\left(4x^2-4x+1\right)-\left(x-1\right)^2\)
\(=\left(2x-1\right)^2-\left(x-1\right)^2\)
\(=\left(2x-1-x+1\right)\left(2x-1+x-1\right)\)
\(=x\left(3x-2\right)\)
= ( 4x^2 + 4x + 1 ) - y^2
= ( 2x + 1 )^2 - y^2
= ( 2x + 1 - y)( 2x + 1 + y)
\(=-\left(4x^2-4x+1\right)=-\left(2x-1\right)^2\)
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