ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}2x-1=a\\\sqrt{x+3}=b\ge0\end{matrix}\right.\) ta được:
\(a+4b=2\sqrt{2a^2+b^2}\)
\(\Rightarrow a^2+8ab+16b^2=8a^2+4b^2\)
\(\Leftrightarrow7a^2-8ab-12b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(7a+6b\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2\sqrt{x+3}=2x-1\\6\sqrt{x+3}=7-14x\end{matrix}\right.\)
\(\Leftrightarrow...\)