\(\left\{{}\begin{matrix}\dfrac{6}{3x-2}-2\sqrt{1-y}=1\\\dfrac{2}{3x-2}+\sqrt{1-y}=2\end{matrix}\right.\) (x \(\ne\) \(\dfrac{2}{3}\); y \(\le\) 1)
Đặt \(\dfrac{1}{3x-2}=a\); \(\sqrt{1-y}=b\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}6a-2b=1\\2a+b=2\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}6a-2b=1\\4a+2b=4\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}10a=5\\4a+2b=4\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=\dfrac{1}{2}\\4\cdot\dfrac{1}{2}+2b=4\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}\dfrac{1}{3x-2}=2\\\sqrt{1-y}=1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2\left(3x-2\right)=1\\1-y=1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{5}{6}\\y=0\end{matrix}\right.\) (TM)
Vậy ...
Chúc bn học tốt!
ĐK:x khác 2/3, y<_1
đặt 1/3x-2=u,căn (1-y) Ta có hệ
6u-2v=1
2u+v=2
sau đó giải hệ và trả ẩn b tự lm nha
