Bài 9:
a: \(A=\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{\sqrt{2}}{2}\)
b: \(B=\dfrac{\sqrt{2}+\sqrt[]{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{8}+2}{\sqrt{2}+\sqrt{3}+2}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt[]{2}\right)}{\sqrt{2}+\sqrt{3}+2}=1+\sqrt{2}\)
Bài 10:
a: \(A=\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)
\(=\dfrac{3\cdot2\sqrt{2}-2\cdot2\sqrt{3}+2\sqrt{5}}{3\cdot3\sqrt{2}-2\cdot3\sqrt{3}+3\sqrt{5}}\)
\(=\dfrac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}=\dfrac{2}{3}\)
b: \(B=\dfrac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)
\(=\dfrac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)}\)
\(=\dfrac{\left(\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{5}-\sqrt{3}\right)}{\left(1-\sqrt{2}\right)\left(2\sqrt{5}-\sqrt{3}\right)}=\dfrac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+1\right)}{2-1}=2+\sqrt{2}-\sqrt{6}-\sqrt{3}\)
