ĐK: \(-\frac{1}{4}\le x\le3\)
Chuyển vế ta có
\(4x+1-2.\sqrt{4x+1}.3+9+3-x-2\sqrt{3-x}+1=0\)
\(\Leftrightarrow\left(\sqrt{4\text{ }x+1}-3\right)^2+\left(\sqrt{3-x}-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{4x+1}=3\\\sqrt{3-x}=1\end{cases}}\)
\(\hept{\begin{cases}4x+1=9\\3-x=1\end{cases}}\Leftrightarrow x=2\)(thỏa ĐK)
k giùm nha