Question

Tìm các số thực x,y biết

\(\dfrac{x+y+7}{z}=\dfrac{y+z+2}{x}=\dfrac{x+z+5}{y}=\dfrac{6}{x+y+z}\)

Answer 1

Lời giải:

Áp dụng TCDTSBN:

$\frac{6}{x+y+z}=\frac{x+y+7}{z}=\frac{y+z+2}{x}=\frac{x+z+5}{y}=\frac{x+y+7+y+z+2+x+z+5}{z+x+y}$

$=\frac{2(x+y+z)+14}{x+y+z}=2+\frac{14}{x+y+z}$

$\Rightarrow \frac{-8}{x+y+z}=2\Rightarrow x+y+z=-4$

Khi đó:

$\frac{x+y+7}{z}=\frac{y+z+2}{x}=\frac{x+z+5}{y}=\frac{6}{x+y+z}=\frac{6}{-4}=\frac{-3}{2}=-1,5$

$\Rightarrow \frac{x+y+z+7}{z}=\frac{x+y+z+2}{x}=\frac{x+y+z+5}{y}=-0,5$

$\Rightarrow \frac{-4+7}{z}=\frac{-4+2}{x}=\frac{-4+5}{y}=-0,5$

$\Rightarrow z=-6; x=4; y=-2$