Question

[ko bt j]

Answer 1

[Akai Haruma]

Lời giải:
$\frac{19}{x+y}+\frac{19}{y+z}+\frac{19}{z+x}=\frac{133}{10}$

$19(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x})=\frac{133}{10}$

$\Rightarrow \frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{133}{10}: 19=\frac{7}{10}(*)$

Lại có:

$\frac{7z}{x+y}+\frac{7x}{y+z}+\frac{7y}{x+z}=\frac{133}{10}$

$\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{x+z}=\frac{133}{10}:7=\frac{19}{10}$

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

$\Rightarrow \frac{x+y+z}{x+y}+\frac{x+y+z}{y+z}+\frac{x+y+z}{x+z}=\frac{49}{10}$

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

Sử dụng kq $(*)$ suy ra:

$x+y+z=\frac{49}{10}: \frac{7}{10}=7$