Câu 8: A
Câu 9: \(\frac{3}{\sqrt5-\sqrt2}+\frac{4}{\sqrt6+\sqrt2}+\frac{1}{\sqrt6+\sqrt5}\)
\(=\frac{3\left(\sqrt5+\sqrt2\right)}{\left(\sqrt5+\sqrt2\right)\left(\sqrt5-\sqrt2\right)}+\frac{4\left(\sqrt6-\sqrt2\right)}{\left(\sqrt6+\sqrt2\right)\left(\sqrt6-\sqrt2\right)}+\frac{1\left(\sqrt6-\sqrt5\right)}{\left(\sqrt6+\sqrt5\right)\left(\sqrt6-\sqrt5\right)}\)
\(=\sqrt5+\sqrt2+\sqrt6-\sqrt2+\sqrt6-\sqrt5=2\sqrt6\)
=>Chọn A
Câu 10: \(\sqrt{\frac49\cdot\frac{1}{81}}=\sqrt{\frac49}\cdot\sqrt{\frac{1}{81}}=\frac23\cdot\frac19=\frac{2}{27}\)
=>Chọn B
Câu 11: \(\sqrt3-\sqrt{27}+\sqrt{48}=\sqrt3-3\sqrt3+4\sqrt3=2\sqrt3\)
=>Chọn C
Câu 12: D
