Question

[Nguyễn Minh Anh]

Giải hệ phương trình

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

Answer 1

[Nguyễn Lê Phước Thịnh]

Đặt \(\sqrt{2x-1}=a;\sqrt{3-y}=b\)

Theo đề, ta có: \(\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{5}{b}=7\\\dfrac{3}{a}-\dfrac{7}{b}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{a}+\dfrac{15}{b}=21\\\dfrac{3}{a}-\dfrac{7}{b}=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{22}{b}=22\\\dfrac{1}{a}+\dfrac{5}{b}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=1\\a=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3-y=1\\2x-1=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=\dfrac{5}{8}\end{matrix}\right.\)