Question

\(\text{1. Tìm x,y,z, biết:}\)

\(x+y=x:y=3\left(x-y\right)\)

Answer 1

Lời giải:

$x+y=\frac{x}{y}$

$y(x+y)=x$

$x(y-1)+y^2=0$

$x(y-1)=-y^2$

Nếu $y=1$ thì $x+1=x$ (vô lý). Do đó $y\neq 1$

$\Rightarrow x=\frac{y^2}{1-y}$.

Khi đó:
$x+y=3(x-y)$

$\Leftrightarrow \frac{y^2}{1-y}+y=\frac{3y^2}{1-y}-3y$

$\Leftrightarrow \frac{y^2}{1-y}=2y$

$\Leftrightarrow y(\frac{y}{1-y}-2)=0$. Rõ ràng $y\neq 0$ nên $\frac{y}{1-y}-2=0$

$\Leftrightarrow y=2(1-y)\Leftrightarrow y=\frac{2}{3}$

$x=\frac{y^2}{1-y}=\frac{4}{3}$