Bài 2:
a: Ta có: \(\sqrt{3x^2+1}=5\)
\(\Leftrightarrow3x^2+1=25\)
\(\Leftrightarrow3x^2=24\)
\(\Leftrightarrow x^2=8\)
hay \(x\in\left\{2\sqrt{2};-2\sqrt{2}\right\}\)
b: Ta có: \(\sqrt{2-3x}=4+\sqrt{6-2\sqrt{5}}-\sqrt{5}\)
\(\Leftrightarrow\sqrt{2-3x}=4+\sqrt{5}-1-\sqrt{5}\)
\(\Leftrightarrow2-3x=9\)
\(\Leftrightarrow3x=11\)
hay \(x=\dfrac{11}{3}\)