Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Xét tích : \(\left[x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)\right]\left(x+y+z\right)\)
=\(x^3\left(z-y\right)+x^2\left(z-y\right)\left(z+y\right)+y^3\left(x-z\right)+y^2\left(x-z\right)\left(x+z\right)\)
\(+z^3\left(y-x\right)+z^2\left(y-x\right)\left(y+x\right)\)
\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)+x^2\left(z^2-y^2\right)+y^2\left(x^2-z^2\right)+z^2\left(y^2-x^2\right)\)
\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)+x^2z^2-x^2y^2+y^2x^2-y^2z^2+z^2y^2-z^2x^2\)
\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)\)
Như vậy:
\(\left[x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)\right]\left(x+y+z\right)\)\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)\)
<=> \(\frac{x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)}{x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)}=x+y+z\)
Ta có: \(\frac{\frac{x^2\left(z-y\right)}{yz}+\frac{y^2\left(x-z\right)}{xz}+\frac{z^2\left(y-x\right)}{xy}}{\frac{x\left(z-y\right)}{yz}+\frac{y\left(x-z\right)}{xz}+\frac{z\left(y-x\right)}{xy}}\)
\(=\frac{\frac{x^3\left(z-y\right)}{xyz}+\frac{y^3\left(x-z\right)}{xyz}+\frac{z^3\left(y-x\right)}{xyz}}{\frac{x^2\left(z-y\right)}{xyz}+\frac{y^2\left(x-z\right)}{xyz}+\frac{z^2\left(y-x\right)}{xyz}}\)
\(=\frac{x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)}{x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)}=x+y+z\)
Ta có : \(x^2+y^2+z^2-xy-yz-zx\)
\(=\frac{1}{2}.\left(2x^2+2y^2+2z^2-2xy-2yz-2zx\right)\)
\(=\frac{\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)}{2}\)
\(=\frac{\left(x-y\right)+\left(y-z\right)^2+\left(z-x\right)^2}{2}\) ( đpcm )
Ta có: \(\frac{x^2-yz}{\left(x+y\right)\left(x+z\right)}=\frac{x^2+xy-xy-yz}{\left(x+y\right)\left(x+z\right)}\)
\(=\frac{x\left(x+y\right)-y\left(x+z\right)}{\left(x+y\right)\left(x+z\right)}\)
\(=\frac{x}{x+z}-\frac{y}{x+y}\)
Tương tự: \(\frac{y^2-xz}{\left(x+y\right)\left(y+z\right)}=\frac{y}{y+z}-\frac{y}{x+y}\)
\(\frac{z^2-xz}{\left(x+z\right)\left(y+z\right)}=\frac{z}{y+z}-\frac{x}{x+z}\)
Do đó: \(A=\frac{x}{x+z}-\frac{y}{x+y}+\frac{y}{y+z}-\frac{x}{x+y}+\frac{z}{y+z}-\frac{x}{x+z}=0\)