Nghiệm của hệ phương trình \(\left\{ \begin{align}  & \sqrt{\frac{1-x}{2y+1}}+\sqrt{\frac{2y+1}{1-x}}=2 \\ & x-y=1 \\\end{align} \right.\) là:

 

Cách giải:

 Đk: 

\(\left\{ \begin{array}{l}\dfrac{{1 - x}}{{2y + 1}} \ge 0\\\dfrac{{2y + 1}}{{1 - x}} \ge 0\\y \ne \dfrac{{ - 1}}{2}\\x \ne 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{1 - x}}{{2y + 1}} > 0\\\dfrac{{2y + 1}}{{1 - x}} > 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}1 - x > 0\\2y + 1 > 0\end{array} \right.\\\left\{ \begin{array}{l}1 - x < 0\\2y + 1 < 0\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x < 1\\y > \dfrac{{ - 1}}{2}\end{array} \right.\\\left\{ \begin{array}{l}x > 1\\y < \dfrac{{ - 1}}{2}\end{array} \right.\end{array} \right..\)

\(\left\{ \begin{array}{l}\sqrt {\dfrac{{1 - x}}{{2y + 1}}} + \sqrt {\dfrac{{2y + 1}}{{1 - x}}} = 2\,\,\,\,\left( 1 \right)\\x - y = 1\,\,\,\,\left( 2 \right)\end{array} \right.\)

từ \(\left( 2 \right)\) suy ra: \(x=1+y\) thay vào \(\left( 1 \right)\) ta có:

\(pt \Leftrightarrow \sqrt {\dfrac{{1 - 1 - y}}{{2y + 1}}}  + \sqrt {\dfrac{{2y + 1}}{{1 - 1 - y}}}  = 2 \Leftrightarrow \sqrt {\dfrac{{ - y}}{{2y + 1}}}  + \sqrt {\dfrac{{2y + 1}}{{ - y}}}  = 2\,\,\,\,\,\,\,\left( 3 \right)\)

Đặt \(\dfrac{-y}{2y+1}=t\left( t\ge 0 \right)\Rightarrow \dfrac{2y+1}{-y}=\dfrac{1}{t}\) khi đó \(\left( 3 \right)\) có dạng:

 \(\dfrac{{ - y}}{{2y + 1}} = 1 \Leftrightarrow 2y + 1 =  - y \Leftrightarrow 3y = 1 \Leftrightarrow y = \dfrac{1}{3} \Rightarrow x = 1 + \dfrac{1}{3} = \dfrac{4}{3}\)

\(\sqrt t  + \sqrt {\dfrac{1}{t}}  = 2 \Leftrightarrow t + 2 + \dfrac{1}{t} = 4 \Leftrightarrow {t^2} - 2t + 1 = 0 \Leftrightarrow {\left( {t - 1} \right)^2} = 0 \Leftrightarrow t = 1\left( {tm} \right)\)

Suy ra: \(\dfrac{-y}{2y+1}=1\Leftrightarrow 2y+1=-y\Leftrightarrow y=\dfrac{1}{3}\,\,\,\left( tm \right)\Rightarrow x=\dfrac{1}{3}+1=\dfrac{4}{3}\,\,\,\left( ktm \right)\)

Vậy hệ phương trình vô nghiệm.