Bài 1:
a) \(\sqrt{6}\cdot\sqrt{3}+\dfrac{2\sqrt{6}}{\sqrt{3}}-\sqrt{32}\)
\(=\sqrt{3}\cdot\sqrt{2}\cdot\sqrt{3}+\dfrac{2\cdot\sqrt{3}\cdot\sqrt{2}}{\sqrt{3}}-\sqrt{4^2\cdot2}\)
\(=3\sqrt{2}+2\sqrt{2}-4\sqrt{2}\)
\(=\left(3+2-4\right)\sqrt{2}\)
\(=\sqrt{2}\)
b) \(\sqrt{23-4\sqrt{15}}-\dfrac{2}{\sqrt{5}+\sqrt{3}}\)
\(=\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}-\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}\)
\(=\sqrt{\left(2\sqrt{5}-\sqrt{3}\right)^2}-\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{2}\)
\(=2\sqrt{5}-\sqrt{3}-\sqrt{5}+\sqrt{3}\)
\(=\sqrt{5}\)
2:
ĐKXĐ: x>=3
\(\sqrt{4x-12}+2\sqrt{9x-27}-\sqrt{36}=\dfrac{2}{5}\sqrt{25x-75}\)
=>\(2\sqrt{x-3}+6\sqrt{x-3}-\dfrac{2}{5}\cdot5\sqrt{x-3}=6\)
=>\(6\sqrt{x-3}=6\)
=>x-3=1
=>x=4