Question

1) Giải hệ phương trình sau:
1/(x + y) + sqrt(y + 2) = 3; - 2/(x + y) + 5sqrt(y + 2) = 1

Answer 1

ĐKXĐ: \(y\ge-2;x+y\ne0\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x+y}=u\\\sqrt{y+2}=v\ge0\end{matrix}\right.\)

Hệ trở thành:

\(\left\{{}\begin{matrix}u+v=3\\-2u+5v=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2u+2v=6\\-2u+5v=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}v=3-v\\7v=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=2\\v=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x+y}=2\\\sqrt{y+2}=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+y=\dfrac{1}{2}\\y+2=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}-y\\y=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-1\end{matrix}\right.\)

Answer 2

Đề thế này đúng ko em: \(\left\{{}\begin{matrix}\dfrac{1}{x+y}+\sqrt{y+2}=3\\\dfrac{-2}{x+y}+5\sqrt{y+2}=1\end{matrix}\right.\)