Question

giải hệ phương trình:

a)\(\left\{{}\begin{matrix}\dfrac{x+1}{x-1}+\dfrac{3y}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)

b)\(\left\{{}\begin{matrix}2\left(x^2-2x\right)+\sqrt{y+1}=0\\3\left(x^2-2x\right)-2\sqrt{y+1}=-7\end{matrix}\right.\)

Answer 1

a) \(\left\{{}\begin{matrix}\dfrac{x+1}{x-1}+\dfrac{3y}{y+2}=7\left(1\right)\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\left(2\right)\end{matrix}\right.\)ĐK: \(x\ne1;y\ne-2\)

(1)\(\Leftrightarrow1+\dfrac{2}{x-1}+3-\dfrac{6}{y+2}=7\Leftrightarrow\dfrac{2}{x-1}-\dfrac{6}{y+2}=3\)

Đặt \(A=\dfrac{1}{x-1};B=\dfrac{1}{y+2}\)

\(\Rightarrow\left\{{}\begin{matrix}2A-6B=3\\2A-5B=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{11}{9}\\y=-1\end{matrix}\right.\)(TM)

Vậy hpt có nghiệm là \(\left(\dfrac{11}{9};-1\right)\).

b)ĐK: \(y\ge-1\)

Đặt \(A=x^2-2x;B=\sqrt{y+1}\left(B\ge0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}2A+B=0\\3A-2B=-7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}A=-1\\B=2\end{matrix}\right.\)(TM)

\(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=1\end{matrix}\right.\)

Vậy hpt có nghiệm là (-1;1);(1;1).