a/ Điều kiện xác định : \(x\ge2\)
\(\sqrt{3x-5}=3+\sqrt{x-2}\)
\(\Leftrightarrow\left(\sqrt{3x-5}\right)^2=\left(3+\sqrt{x-2}\right)^2\)
\(\Leftrightarrow3x-5=9+x-2+6\sqrt{x-2}\)
\(\Leftrightarrow x-6=3\sqrt{x-2}\)
\(\Leftrightarrow\left(x-6\right)^2=9\left(x-2\right)\)
\(\Leftrightarrow x^2-12x+36=9x-18\)
\(\Leftrightarrow x^2-21x+54=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-18\right)=0\Leftrightarrow\orbr{\begin{cases}x=3\\x=18\end{cases}}\) (TM)
Vậy..........................................................
b/ ĐKXĐ : \(x\ge\frac{2}{5}\)
\(\sqrt{25x^2-4}=2\sqrt{5x-2}\)
\(\Leftrightarrow25x^2-4=4\left(5x-2\right)\) (bình phương hai vế )
\(\Leftrightarrow25x^2-20x+4=0\)
\(\Leftrightarrow\left(5x-2\right)^2=0\Leftrightarrow x=\frac{2}{5}\) (TM)
Vậy ................................................
