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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)
(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
Ta có:\(2\left(x-y\right)\left(z-y\right)+2\left(y-z\right)\left(z-x\right)+2\left(y-z\right)\left(x-z\right)\)
\(=2\left[\left(x-y\right)\left(z-y\right)+\left(y-x\right)\left(z-x\right)+\left(y-z\right)\left(x-z\right)\right]\)
\(=2\left[xz-xy-yz+y^2+yz-xy-zx+x^2+yx-yz-zx+z^2\right]\)
\(=2\left[-xz-xy-yz+x^2+y^2+z^2\right]\)
\(=x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2\)
\(=\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\)
a) Ta có:
(x+y+z)(x-y-z) = x^2 -xy -xz +yx- y^2 -yz+zx -zy -z^2
=x^2 - y^2 - 2yz - z^2.
b) Ta có: (x-y+z)(x+y+z) = x^2 +xy+xz -yx-y^2 -yz +zx+zy +z^2
=x^2 +2xz- y^2 +z^2.
c) Ta có: -16 + (x-3)^2 = -16 + ( x^2-6x+9)
= -16 + x^2 - 6x + 9
= x^2 - 6x - 7.
\(a,\left(x+y+z\right)\left(x-y-z\right)\)
\(=x\left(x-y-z\right)+y\left(x-y-z\right)+z\left(x-y-z\right)\)
\(=x^2-xy-xz+xy-y^2-yz+xz-yz-z^2\)
\(=x^2-y^2-2yz-z^2\)
\(b,\left(x-y+z\right)\left(x+y+z\right)\)
\(=x\left(x+y+z\right)-y\left(x+y+z\right)+z\left(x+y+z\right)\)
\(=x^2+xy+xz-xy-y^2-yz+xz+yz+z^2\)
\(=x^2+2xz-y^2+z^2\)
\(c,-16+\left(x-3\right)^2\)
\(=-16+x^2-6x+9\)
\(=x^2-6x-7\)
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, (x + y + z)(x - y - z)
= x^2 - xy - xz + xy - y^2 - zy + zx - zy - z^2
= x^2 + y^2 + z^2 + (xy - xy) + (xz - xz) - (zy + zy)
= x^2 + y^2 + z^2 - 2zy
b, (x - y + z)(x + y + z)
= x^2 + xy + xz - xy - y^2 - zy + zx + zy + z^2
= x^2 + y^2 + z^2 + (xy - xy) + xz + xz + (zy - zy)
= x^2 + y^2 + z^2 + 2zx
( x + y + z +t )( x + y - z - t)
= ( x + y)^2 - ( z + t)^2
= x^2 + 2xy + y^2 - z^2 - 2zt - 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 \(S=\left\{1;-7\right\}\)
\(x^2+6x-7=0\\ \Leftrightarrow x^2+7x-x-7=0\\ \Leftrightarrow\left(x^2+7x\right)-\left(x+7\right)=0\\ \Leftrightarrow x\left(x+7\right)-\left(x+7\right)=0\\ \Leftrightarrow\left(x+7\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-7\\x=1\end{matrix}\right.\)