BT

GHPT
$\begin{cases}x^{2}+x+\frac{1}{y}(1+\frac{1}{y})= 4\\ \frac{x^{2}}{y}+\frac{x}{y^{2}}+\frac{1}{y^{3}}=4-x^{3} \end{cases}$ 

Phương trình đã cho $\Leftrightarrow \begin{cases}(x+\frac{1}{y})^{2}+(x+\frac{1}{y})-\frac{2x}{y}=4 \\ (x+\frac{1}{y})^{3}-\frac{2x}{y}(x+\frac{1}{y})=4 \end{cases}$
 

Đặt:   $\begin{cases}x+\frac{1}{y}= u\\ \frac{2x}{y}=v \end{cases}\Rightarrow \begin{cases}u^{2}+u-v=4 \\ u^{3}-uv=4 \end{cases}\Leftrightarrow \begin{cases}v=u^{2}+u-4 \\ u=2 \end{cases}\Leftrightarrow u=v=2$

$\begin{cases}x+\frac{1}{y}= 2\\ x=y \end{cases}\Leftrightarrow x=y=1$
Mình cũng ra kết quả thế này –  nguyenthihakhanhtb 20-02-13 03:41 PM
$Đặt t=\frac{1}{2}\Rightarrow$
\begin{cases}xt(x+t)+(x+t)((x+t)^2-3xt)=4 \\ (x+t)-xt=2 \end{cases}
$Đặt u=x+t;v=xt \Rightarrow$
\begin{cases}uv+u^3-3v=4 \\ u-v=2 \end{cases}
$\Leftrightarrow  u^3+u^2-5u-2=0 \Leftrightarrow u=2 \rightarrow v=0 thay lại thấy pt vô n\vee u= \frac{-3\pm \sqrt{5}}{2} thay tở lại.... $
Vote up nều câu trả lời đúng –  xxxjimmy 19-02-13 06:18 PM

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