1) Ta có: \(\sqrt{11}-\sqrt{2}=\frac{\left(\sqrt{11}-\sqrt{2}\right)\left(\sqrt{11}+\sqrt{2}\right)}{\sqrt{11}+\sqrt{2}}\)
\(=\frac{11-2}{\sqrt{11}+\sqrt{2}}=\frac{9}{\sqrt{11}+\sqrt{2}}\)
Ta có: \(\sqrt{14}-\sqrt{5}=\frac{\left(\sqrt{14}-\sqrt{5}\right)\left(\sqrt{14}+\sqrt{5}\right)}{\sqrt{14}+\sqrt{5}}\)
\(=\frac{14-5}{\sqrt{14}+\sqrt{5}}=\frac{9}{\sqrt{14}+\sqrt{5}}\)
Ta có: \(\left(\sqrt{11}+\sqrt{2}\right)^2=11+2\cdot\sqrt{22}+2=13+2\sqrt{22}\)
\(\left(\sqrt{14}+\sqrt{5}\right)^2=14+2\cdot\sqrt{70}+5=19+2\sqrt{70}=13+2\sqrt{70}+6\)
Ta có: \(2\sqrt{22}< 2\sqrt{70}\)
\(\Leftrightarrow13+2\sqrt{22}< 13+2\sqrt{70}\)
mà \(13+2\sqrt{70}< 19+2\sqrt{70}\)
nên \(13+2\sqrt{22}< 19+2\sqrt{70}\)
\(\Leftrightarrow\left(\sqrt{11}+\sqrt{2}\right)^2< \left(\sqrt{14}+\sqrt{5}\right)^{^2}\)
\(\Leftrightarrow\sqrt{11}+\sqrt{2}< \sqrt{14}+\sqrt{5}\)
\(\Leftrightarrow\frac{9}{\sqrt{11}+\sqrt{2}}>\frac{9}{\sqrt{14}+\sqrt{5}}\)
hay \(\sqrt{11}-\sqrt{2}>\sqrt{14}-\sqrt{5}\)
2) Ta có: \(\left(\sqrt{5}+\sqrt{7}\right)^2=5+2\cdot\sqrt{35}+7=12+2\sqrt{35}=12+\sqrt{140}\)
\(\left(2\sqrt{6}\right)^2=4\cdot6=24=12+12=12+\sqrt{144}\)
mà \(12+\sqrt{140}< 12+\sqrt{144}\)
nên \(\left(\sqrt{5}+\sqrt{7}\right)^2< \left(2\sqrt{6}\right)^2\)
hay \(\sqrt{5}+\sqrt{7}< 2\sqrt{6}\)