\(M=\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x}=\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{z+x}+1-3..\)
= \(\frac{x+y+z}{x+y}+\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}-3.\)
= \(\left(x+y+z\right).\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)-3.\)
= \(2010.\frac{1}{2018}-3=\frac{-2022}{1009}.\)
Ta có:\(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}=\frac{1}{2018}\)
Nhân cả hai vế với (x+y+z) ta có:
\(\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}+\frac{x+y+z}{x+y}=\frac{x+y+z}{2018}\)
\(\Rightarrow1+\frac{x}{y+z}+1+\frac{y}{z+x}+1+\frac{z}{x+y}=\frac{2010}{2018}\)
\(\Rightarrow3+M=\frac{1005}{1009}\)
\(\Rightarrow M=\frac{1005}{1009}-3\)
\(\Rightarrow M=\frac{-2022}{1009}\)
Có \(\frac{1}{z+y}+\frac{1}{z+x}+\frac{1}{x+y}=\frac{1}{2018}\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{z+y}+\frac{1}{z+x}+\frac{1}{x+y}\right)=\frac{x+y+z}{2018}\)
\(\Leftrightarrow\frac{x+y+z}{y+z}+\frac{x+y+z}{y+z}+\frac{x+y+z}{x+y}=\frac{2010}{2018}\) \(\Leftrightarrow1+\frac{x}{y+z}+1+\frac{y}{z+x}+1+\frac{z}{x+y}=\frac{2010}{2018}\)
\(\Leftrightarrow\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=\frac{2010}{2008}-3\)
\(\Leftrightarrow\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=\frac{-2022}{1009}\)
Bài làm của cô Vân có 1 chỗ sai,em xin sửa lại
\(\Leftrightarrow\frac{x+y+z}{y+z}+\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}\)
Cách mình hơi khác
\(\frac{1}{z+y}+\frac{1}{z+x}+\frac{1}{x+y}=\frac{\left(z+y\right)+\left(z+x\right)+\left(x+y\right)}{\left(z+y\right)\left(z+x\right)\left(x+y\right)}\)
\(\Rightarrow\frac{2\left(x+y+z\right)}{\left(z+y\right)\left(z+x\right)\left(y+x\right)}\)
\(\Rightarrow\frac{x+y+z}{x+y}+\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}=\frac{2010}{2018}\)
\(\Rightarrow\left(1+1+1\right)+\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=\frac{2010}{2018}\)
\(\Rightarrow\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x}=\frac{2010}{2018}-3\)
\(\Rightarrow M=\frac{-2022}{1009}\)
