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a, \(\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)^2}\) +\(\frac{1}{\left(\sqrt{3}-\sqrt{2}\right)^2}\) =\(\frac{\left(\sqrt{3}+\sqrt{2}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\frac{10}{1}=10\)
mấy câu còn lại bạn tự làm nốt nhé mk ban rồi
a) \(\sqrt{17}-4\) b) \(\sqrt{3}\) c) \(\frac{\sqrt{2}}{2}\) d)\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\) e) \(x-\sqrt{5}\)
f) \(4+2\sqrt{3}\) g) \(3+2\sqrt{2}\) h) \(x+\sqrt{x}+1\) i) \(\frac{3\sqrt{5}-\sqrt{15}}{10}\)
k) \(\sqrt{5}+\sqrt{6}\) i) 5 h) 0 l) \(\sqrt{5}+\sqrt{3}\) m) \(\frac{20\sqrt{3}}{3}\) d) 0
a)
\(\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}=\frac{2(\sqrt{6}+2+\sqrt{6}-2)}{(\sqrt{6}-2)(\sqrt{6}+2)}+\frac{5\sqrt{6}}{6}\)
\(=\frac{4\sqrt{6}}{6-2^2}+\frac{5\sqrt{6}}{6}=2\sqrt{6}+\frac{5\sqrt{6}}{6}=\frac{17\sqrt{6}}{6}\)
b)
\(\frac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}-\frac{1}{\sqrt{3}+\sqrt{2}+\sqrt{5}}=\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}-(\sqrt{3}+\sqrt{2}-\sqrt{5})}{(\sqrt{3}+\sqrt{2}-\sqrt{5})(\sqrt{3}+\sqrt{2}+\sqrt{5})}\)
\(=\frac{2\sqrt{5}}{(\sqrt{3}+\sqrt{2})^2-5}=\frac{2\sqrt{5}}{5+2\sqrt{6}-5}=\sqrt{\frac{5}{6}}\)
c)
\(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right):\frac{1}{\sqrt{5}-\sqrt{2}}\)
\(=\left[\frac{\sqrt{2}(\sqrt{3}-1)}{1-\sqrt{3}}-\sqrt{5}\right].(\sqrt{5}-\sqrt{2})\)
\(=(-\sqrt{2}-\sqrt{5})(\sqrt{5}-\sqrt{2})=-(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})\)
\(=-(5-2)=-3\)
d)
\(\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)
\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{1}{4}+\frac{2}{2\sqrt{6}}+\frac{1}{6}}\)
\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{(\frac{1}{2}-\frac{1}{\sqrt{6}})^2}\)
\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}(\frac{1}{2}-\frac{1}{\sqrt{6}})\)
\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{2\sqrt{3}}-\frac{1}{3\sqrt{2}}=\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2}\)