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a ) \(\frac{1}{\left(x-y\right)\left(y-z\right)}+\frac{1}{\left(y-z\right)\left(z-x\right)}+\frac{1}{\left(z-x\right)\left(x-y\right)}\)
= \(\frac{z-x}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{x-y}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
= \(\frac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)
b ) \(\frac{4}{\left(y-x\right)\left(z-x\right)}+\frac{3}{\left(y-x\right)\left(y-z\right)}+\frac{3}{\left(y-z\right)\left(x-z\right)}\)
= \(\frac{-4}{\left(y-x\right)\left(x-z\right)}+\frac{3}{\left(y-x\right)\left(y-z\right)}+\frac{3}{\left(y-z\right)\left(x-z\right)}\)
= \(\frac{-4\left(y-z\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}+\frac{3\left(x-z\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}+\frac{3\left(y-x\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}\)
= \(\frac{-4y+4z+3x-3z+3y-3x}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}=\frac{z-y}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}\)
= \(\frac{-\left(y-x\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}=\frac{-1}{\left(x-z\right)\left(y-z\right)}=\frac{1}{\left(x-z\right)\left(x-y\right)}\)
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cậu vào đường link này sẽ rõ:http://olm.vn/hoi-dap/question/794605.html
\(\frac{1}{\left(x-y\right)\left(y-z\right)}+\frac{1}{\left(y-z\right)\left(z-x\right)}+\frac{1}{\left(z-x\right)\left(x-y\right)}\)
\(=\frac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=0\)
Ta có : \(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1\) \(\Rightarrow\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+\frac{xy+xz}{y+z}+\frac{y^2}{z+x}+\frac{xy+yz}{z+x}+\frac{z^2}{x+y}+\frac{zx+zy}{x+y}\)\(=x+y+z\)
\(\Rightarrow P+\frac{x\left(y+z\right)}{y+z}+\frac{y\left(x+z\right)}{x+z}+\frac{z\left(x+y\right)}{x+y}=x+y+z\)
\(\Rightarrow P+x+y+z=x+y+z\Rightarrow P=0\)
Vậy P = 0