Bài 2:
a: Ta có: \(\sqrt{36x^2-12x+1}=5\)
\(\Leftrightarrow\left|6x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}6x-1=5\\6x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b: Ta có: \(\sqrt{x-5}+2\sqrt{4x-20}-\dfrac{1}{3}\sqrt{9x-45}=12\)
\(\Leftrightarrow\sqrt{x-5}-\sqrt{x-5}+4\sqrt{x-5}=12\)
\(\Leftrightarrow x-5=9\)
hay x=14
\(1,\\ a,=\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\\ =\left(\sqrt{5}-\sqrt{3}\right)-\left(2\sqrt{5}-\sqrt{3}\right)=-\sqrt{5}\\ b,=\left(\dfrac{3\sqrt{6}}{2}+\dfrac{2\sqrt{6}}{3}-2\sqrt{6}\right)\left(\sqrt{6}-2\sqrt{3}-\sqrt{6}\right)\\ =\dfrac{9\sqrt{6}+4\sqrt{6}-12\sqrt{6}}{6}\left(-2\sqrt{3}\right)\\ =\dfrac{\sqrt{6}\left(-2\sqrt{3}\right)}{6}=\dfrac{-6\sqrt{2}}{6}=-\sqrt{2}\)
\(c,=\dfrac{4\left(\sqrt{3}-1\right)}{2}-\dfrac{5\left(2+\sqrt{3}\right)}{-1}+\dfrac{6\left(3+\sqrt{3}\right)}{-6}\\ =2\sqrt{3}-2+10+5\sqrt{3}-3-\sqrt{3}\\ =6\sqrt{3}-5\)
