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\(\dfrac{x-y}{z^2+1}=\dfrac{x-y}{z^2+xy+yz+zx}=\dfrac{x-y}{z\left(z+y\right)+x\left(z+y\right)}=\dfrac{x-y}{\left(x+z\right)\left(z+y\right)}\)
Tương tự: \(\dfrac{y-z}{x^2+1}=\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}\);\(\dfrac{z-x}{y^2+1}=\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)
Cộng vế với vế \(\Rightarrow VT=\dfrac{x-y}{\left(x+z\right)\left(y+z\right)}+\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}+\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)+\left(y-z\right)\left(y+z\right)+\left(z-x\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\dfrac{x^2-y^2+y^2-z^2+z^2-x^2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=0\)(đpcm)
\(\frac{x^2-yz}{yz}+1+\frac{y^2-zx}{zx}+1+\frac{z^2-xy}{xy}+1=3\Leftrightarrow\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=3\)
\(\Leftrightarrow\frac{1}{xyz}\left(x^3+y^3+z^3\right)=3\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)
Tới đây bạn thay vào nhé :)
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=0\Rightarrow\frac{x+y+z}{xyz}=0\Rightarrow x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)
\(N=\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=\frac{x^3+y^3+z^3}{xyz}=\frac{3xyz}{xyz}=3\)
\(M=\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\)
\(=\frac{x^2y^2+y^2z^2+z^2x^2}{xyz}\)
\(=\frac{\left(xy+yz+zx\right)^2-2x^2yz-2xyz^2-2x^2yz}{xyz}\)
\(=\frac{0-2xyz\left(x+y+z\right)}{xyz}\)
\(=0-2\left(x+y+z\right)\)
\(=0-2.\left(-1\right)=0-\left(-2\right)=2\)
Chúc bạn học tốt.
Vì xy + yz + zx = 1 ta có :
\(\frac{x-y}{z^2+1}+\frac{y-z}{x^2+1}+\frac{z-x}{y^2+1}=\frac{x-y}{z^2+xy+yz+zx}+\frac{y-z}{x^2+xy+yz+zx}+\frac{z-x}{y^2+xy+yz+zx}\)
\(=\frac{x-y}{\left(y+z\right)\left(z+x\right)}+\frac{y-z}{\left(x+y\right)\left(x+z\right)}+\frac{z-x}{\left(y+z\right)\left(x+y\right)}\)
\(=\frac{\left(x-y\right)\left(x+y\right)+\left(y-z\right)\left(y+z\right)+\left(x+z\right)\left(z-x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\frac{x^2-y^2+y^2-z^2+z^2-x^2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{0}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=0\)(ĐPCM)
Ta có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)^3=-\frac{1}{z^3}\)
\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+3\cdot\frac{1}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{z^3}=0\)
\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=-3\cdot\frac{1}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)=-3\cdot\frac{1}{xy}\cdot\left(-\frac{1}{z}\right)=\frac{3}{xyz}\)
Khi đó có : \(P=\frac{yz}{x^2}+\frac{zx}{y^2}+\frac{xy}{z^2}=xyz.\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz\cdot\frac{3}{xyz}=3\)
GT \(\Leftrightarrow xy+yz+zx=0\). Khi đó: \(\left(xy\right)^3+\left(yz\right)^3+\left(zx\right)^3=3.xy.yz.zx=3x^2y^2z^2\).
Do đó: \(P=\frac{\left(xy\right)^3+\left(yz\right)^3+\left(zx\right)^3}{x^2y^2z^2}=3\)