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\(5\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}-\sqrt{\frac{5}{2}}\right)^2+\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}-\sqrt{\frac{3}{2}}\right)^2\)
\(=\frac{5}{2}\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}-\sqrt{5}\right)^2+\frac{1}{2}\left(\sqrt{4-2\sqrt{3}}+\sqrt{6+2\sqrt{5}}-\sqrt{3}\right)^2\)
\(=\frac{5}{2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{5}\right)^2+\frac{1}{2}\left(\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{3}\right)^2\)
\(=\frac{5}{2}\left(\sqrt{3}+1+\sqrt{5}-1-\sqrt{5}\right)^2+\frac{1}{2}\left(\sqrt{3}-1+\sqrt{5}+1-\sqrt{3}\right)^2\)
\(=\frac{5}{2}\left(\sqrt{3}\right)^2+\frac{1}{2}\left(\sqrt{5}\right)^2=\frac{15}{2}+\frac{5}{2}=\frac{20}{2}=10\)
\(P=\frac{3\sqrt{2}+\sqrt{10}}{2\sqrt{5}+\sqrt{6+2\sqrt{5}}}-\frac{3\sqrt{2}-\sqrt{10}}{2\sqrt{5}+\sqrt{6-2\sqrt{5}}}=\frac{3\sqrt{2}+\sqrt{10}}{2\sqrt{5}+\sqrt{5}+1}-\frac{3\sqrt{2}-\sqrt{10}}{2\sqrt{5}+\sqrt{5}-1}\)
\(=\frac{3\sqrt{2}+\sqrt{10}}{3\sqrt{5}+1}-\frac{3\sqrt{2}-\sqrt{10}}{3\sqrt{5}-1}=\frac{9\sqrt{10}-3\sqrt{2}+15\sqrt{2}-\sqrt{10}-9\sqrt{10}-3\sqrt{2}+15\sqrt{2}+\sqrt{10}}{44}\)
\(=\frac{24\sqrt{2}}{44}=\frac{6\sqrt{2}}{11}.\)
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
\(A=\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{3}-1}\)
\(=\frac{\sqrt{3}-1}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\frac{\sqrt{3}+1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{3}-1+\sqrt{3}+1}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\sqrt{3}}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\sqrt{3}}{3-1}\)
\(=\frac{2\sqrt{3}}{2}\)
\(=\sqrt{3}\)
\(B=\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\)
\(=\frac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}\left(\sqrt{5}-1\right)}+\frac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}\left(\sqrt{5}+1\right)}\)
\(=\frac{\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)}+\frac{\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)}\)
\(=\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}+\frac{\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
\(=\frac{5+2\sqrt{5}+1+5-2\sqrt{5}+1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(=\frac{12}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(=\frac{12}{5-1}\)
\(=\frac{12}{4}\)
\(=3\)
a) Ta có: \(\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\frac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
\(=\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\frac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{5}-\frac{\sqrt{5}}{2}=\frac{2\sqrt{5}}{2}-\frac{\sqrt{5}}{2}=\frac{\sqrt{5}}{2}\)
b) Ta có: \(\frac{5+\sqrt{5}}{\sqrt{10}+\sqrt{2}}\)
\(=\frac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}\)
\(=\frac{\sqrt{5}}{\sqrt{2}}\)
\(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\)
\(=\frac{\left(5+\sqrt{5}\right)\left(5+\sqrt{5}\right)}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)\(+\frac{\left(5-\sqrt{5}\right)\left(5-\sqrt{5}\right)}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
\(=\frac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{25-5}\)
\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{20}\)
\(=\frac{60}{20}=30\)
Chúc bạn học tốt!