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\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(\left(x-1\right)^2-2\left(x-1\right)\left(2x+1\right)+\left(2x+1\right)^2\)
\(=\left(x-1-2x-1\right)^2=\left(-x-2\right)^2=\left(x+2\right)^2\)
x(x+2)(x^2+2x+2)+1
(x^2+2x)(x^2+2x+2)+1
dat x^2+2x=a
=> a(a+2)+1
=a^2+2a+1
=(a+1)^2
=(x^2+2x+1)^2
=(x+1)^4
x(x+2)(x^2+2x+2)+1
(x^2+2x)(x^2+2x+2)+1
dat x^2+2x=a
=> a(a+2)+1
=a^2+2a+1
=(a+1)^2
=(x^2+2x+1)^2
=(x+1)^4
Ta có \(\left(1+2x\right)\left(1-2x\right)-x\left(x+2\right)\left(x-2\right)\)
\(=1-4x^2-x\left(x^2-4\right)=1-4x^2-x^3+4x\)
\(=\left(1-x^4\right)+4x\left(1-x\right)=\left(1-x\right)\left(x^2+x+1\right)+4x\left(1-x\right)\)
\(=\left(1-x\right)\left(x^2+5x+1\right)\)
\(\left(x^2-x+2\right)\left(x-1\right)-x^2\left(x-1\right)^2+\left(2x+1\right)\left(x-1\right)^3\)
\(=\left(x-1\right)\left[x^2-x+2-x^2\left(x-1\right)+\left(2x+1\right)\left(x^2-2x+1\right)\right]\)
\(=\left(x-1\right)\left(x^2-x+2-x^3+x^2+2x^3-4x^2+2x+x^2-2x+1\right)\)
\(=\left(x-1\right)\left(x^3-x^2-x+3\right)\)
\(2\left(x^2+x+1\right)^2-\left(2x+1\right)^2-\left(x^2+2x\right)^2\)
\(=2\left(x^4+2x^3+3x^2+2x+1\right)-4x^2-4x-1-x^4-4x^3-4x^2\)
\(=2x^4+4x^3+6x^2+4x+2-4x^2-4x-1-x^4-4x^3-4x^2\)
\(=x^4-2x^2+1\)
\(=\left(x^2-1\right)^2\)
\(=\left[\left(x-1\right)\left(x+1\right)\right]^2\)
\(=\left(x-1\right)^2\left(x+1\right)^2\)
Chúc bạn học tốt.
x(x+2)(x^2+2x+2)+1=(x^2+2x+1-1)(x^2+2x+1+1)+1
=( x^2+2x+1)^2-1+1
=(x^2+2x+1)^2