Question

[린 린]

\(\text{Giải hệ pt}\)\(\hept{\begin{cases}\sqrt{x+1}-\frac{2}{2x+y}=-1\\\sqrt{4x+4}+\frac{3}{2x+y}=5\end{cases}}\)

Answer 1

[T.Anh 2K7(siêu quậy)(тoáɴ๖ۣۜнọc)]

\(ĐKXĐ:x\ge-1;2x+y\ne0\)

Ta có:\(\sqrt{x+1}-\frac{2}{2x+y}=-1\Rightarrow3\sqrt{x+1}-\frac{6}{2x+y}=-3\left(1\right)\)

\(\sqrt{4x+4}+\frac{3}{2x+y}=5\Rightarrow2\sqrt{4\left(x+1\right)}+\frac{6}{2x+y}=10\Rightarrow4\sqrt{x+1}+\frac{6}{2x+y}=10\left(2\right)\)

Lấy (1) cộng (2) ta được:

\(\Rightarrow4\sqrt{x+1}+3\sqrt{x+1}=7\Rightarrow7\sqrt{x+1}=7\Rightarrow\sqrt{x+1}=1\Rightarrow x+1=1\Rightarrow x=0\left(TM\right)\)

Khi đó ta có:\(\Rightarrow\sqrt{0+1}-\frac{2}{2.0+y}=-1\Rightarrow1-\frac{2}{y}=-1\Rightarrow\frac{2}{y}=2\Rightarrow y=1\)

                 Vậy \(x,y\in\left\{0;1\right\}\)