Question

Tính tổng M=x+y+z, biết \(\dfrac{19}{x+y}+\dfrac{19}{y+z}+\dfrac{19}{z+x}=\dfrac{7x}{y+z}+\dfrac{7y}{z+x}+\dfrac{7z}{x+y}=\dfrac{133}{10}\)

Answer 1

Lời giải:

Ta có:

\(\frac{19}{x+y}+\frac{19}{y+z}+\frac{19}{z+x}=\frac{133}{10}\)

\(\Leftrightarrow \frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{7}{10}(*)\)

Và: \(\frac{7x}{y+z}+\frac{7y}{z+x}+\frac{7z}{x+y}=\frac{133}{10}\)

\(\Leftrightarrow \frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=\frac{19}{10}\)

\(\Rightarrow \frac{x}{y+z}+1+\frac{y}{z+x}+1+\frac{z}{x+y}+1=\frac{49}{10}\)

\(\Leftrightarrow \frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}+\frac{x+y+z}{x+y}=\frac{49}{10}\)

\(\Leftrightarrow (x+y+z)\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right)=\frac{49}{10}(**)\)

Từ \((*); (**)\Rightarrow x+y+z=\frac{49}{10}:\frac{7}{10}=7\)

Vậy $M=7$