(x+y+z)(x+z-y)(x+y-z)(y+z-x)

=[(x+y)^2-z^2]*[(x+z-y)(y+z-x)]

=[(x+y)^2-z^2][y^2-(x+z)^2]

=(x^2+2xy+y^2-z^2][y^2-x^2-2xz-z^2]

=x^2y^2-x^4-2x^3z-x^2z^2+2xy^3-2x^3y-4x^2yz-2xyz^2+y^4-y^2x^2-2xy^2z-z^2y^2-y^2z^2+x^2z^2+2xz^3+z^4

`(x+y+z+t)(x+y-z-t)`

`=[(x+y)+(z+t)][(x+y)-(z+t)]`

`=(x+y)^2-(z-t)^2`

`=(x+y)^2+[-(z-t)^2]`

\(\left(x+y+z+t\right)\left(x+y+z-t\right)=\left(x+y+z\right)^2-t^2\)

\(x^2+6x-7=0\\ \Leftrightarrow x^2-x+7x-7=0\\ \Leftrightarrow x\left(x-1\right)+7\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)

Vậy \(S=\left\{1;-7\right\}\)

\(x^2+6x-7=0\\ \Leftrightarrow x^2+7x-x-7=0\\ \Leftrightarrow\left(x^2+7x\right)-\left(x+7\right)=0\\ \Leftrightarrow x\left(x+7\right)-\left(x+7\right)=0\\ \Leftrightarrow\left(x+7\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-7\\x=1\end{matrix}\right.\)

\(x^2+6x-7=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)

Vậy \(S=\left\{1;-7\right\}\)

Ta có:\(\left(x-y+z\right)\left(x+y+z\right)=\left[\left(x+z\right)-y\right]\left[\left(x+z\right)+y\right]\)

\(=\left(x+z\right)^2-y^2=x^2-2xz+z^2-y^2\)

Xong rồi đấy,chúc bạn học tốt

a: \(\left(a^2+2a+3\right)\left(a^2-2a-3\right)\)

\(=\left[a^2+\left(2a+3\right)\right]\left[a^2-\left(2a+3\right)\right]\)

\(=\left(a^2\right)^2-\left(2a+3\right)^2\)

\(=a^4-\left(2a+3\right)^2\)

b: \(\left(-a^2-2a+3\right)^2\)

\(=\left(a^2+2a-3\right)^2\)

\(=a^4+4a^2+9+4a^3-18a-6a^2\)

\(=a^4+4a^3-2a^2-18a+9\)

c: \(\left(x-y-z\right)^2\)

\(=x^2-2x\left(y+z\right)+\left(y+z\right)^2\)

\(=x^2-2xy-2xz+y^2+2yz+z^2\)

d: \(\left(x+y+z\right)\left(x-y-z\right)\)

\(=x^2-\left(y+z\right)^2\)

\(=x^2-y^2-2yz-z^2\)

\(\left(x+y+z\right)\left(x-y-z\right)\)

\(=x^2-xy-xz+xy-y^2-yz+xz-yz-z^2\)

\(=x^2-y^2-z^2-2yz\)

2:

-8x^6-12x^4y-6x^2y^2-y^3

=-(8x^6+12x^4y+6x^2y^2+y^3)

=-(2x^2+y)^3

3:

=(1/3)^2-(2x-y)^2

=(1/3-2x+y)(1/3+2x-y)

Ta có:\(2\left(x-y\right)\left(z-y\right)+2\left(y-z\right)\left(z-x\right)+2\left(y-z\right)\left(x-z\right)\)

\(=2\left[\left(x-y\right)\left(z-y\right)+\left(y-x\right)\left(z-x\right)+\left(y-z\right)\left(x-z\right)\right]\)

\(=2\left[xz-xy-yz+y^2+yz-xy-zx+x^2+yx-yz-zx+z^2\right]\)

\(=2\left[-xz-xy-yz+x^2+y^2+z^2\right]\)

\(=x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2\)

\(=\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\)