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\(x^5+x^4+x^3+x^2+x+1\)
\(=\left(x^5+x^4+x^3\right)+\left(x^2+x+1\right)\)
\(=x^3\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^3+1\right)\left(x^2+x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
Ta có: \(x^5+x^4+x^3+x^2+x+1\)
\(=x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^4+x^2+1\right)\)
\(=\left(x+1\right)\left(x^4+2x^2+1-x^2\right)\)
\(=\left(x+1\right)\left\lbrack\left(x^2+1\right)^2-x^2\right\rbrack=\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)

Câu 1 : x2-y2+2yz-z2=-(y2-2yz+z2-x2) Câu 2: x2-2xy+y2-xz+yz=(x2-2xy+y2)-xz+yz
=-(y-z)2 -x2 =(x-y)2-z(x-y)
=-(y-z-x)(y-z+x) =(x-y)(x-y-z)

ta có :
\(\left(2x+1\right)^2-\left(x-1\right)^2=\left(2x+1+x-1\right)\left(2x+1-x+1\right)=3x\left(x+2\right)\)

x2 - 2xy - 4z2 + y2
= (x2 - 2xy+y2) - (2z)2
= (x-y)2- (2z)2
= (x-y-2z)(x-y+2z)
x2-2xy-4z2+y2
=(x2-2xy+y2)-(2z)2
=(x-y)2-(2z)2
=(x-y-2z)(x-y+2z)
x^2+2x+1−y^2
(a+b)^2=a^2+2ab+b^2
x^2+2x+1=(x+1)^2
(x+1)^2−y^2
a^2−b^2=(a−b)(a+b)
(x+1)^2−y^2=(x+1−y)(x+1+y)
=(x−y+1)(x+y+1)
\(x^2\) + 2\(x\) + 1 - y\(^2\)
= (\(x^2\) + 2.\(x\).1 + 1\(^2\)) - y\(^2\)
= (\(x+1\))\(^2\) - y\(^2\)
= (\(x+1-y\))(\(x+1+y\))