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Đặt: \(A=\frac{\sqrt{7+\sqrt{5}}+\sqrt{7-\sqrt{5}}}{\sqrt{7+2\sqrt{11}}}\)\(>\)\(0\)
=> \(A^2=\frac{7+\sqrt{5}+2.\sqrt{\left(7+\sqrt{5}\right)\left(7-\sqrt{5}\right)}+7-\sqrt{5}}{7+2\sqrt{11}}\)
\(=\frac{14+4\sqrt{11}}{7+2\sqrt{11}}\)
\(=\frac{2\left(7+2\sqrt{11}\right)}{7+2\sqrt{11}}=2\)
=> \(A=\sqrt{2}\)
\(D=\sqrt{2}-\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\sqrt{2}-\left(\sqrt{2}-1\right)=1\)
a) \(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{49-48}=14\)
b) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)
a) đặt \(A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
nhân cả hai vế với \(\sqrt{2}\), ta được:
\(\sqrt{2}A=\sqrt{2}\sqrt{4-\sqrt{7}}-\sqrt{2}\sqrt{4+\sqrt{7}}\)
\(=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(1+ \sqrt{7}\right)^2}\)
\(=\left|1-\sqrt{7}\right|-\left|1+\sqrt{7}\right|\)
\(=\sqrt{7}-1-\sqrt{7}-1\)
\(=-2\)
\(\Rightarrow A=-\frac{2}{\sqrt{2}}=-\sqrt{2}\)
\(\frac{2}{\sqrt{7}-5}-\frac{2}{\sqrt{7}+5}=\frac{2\sqrt{7}+10}{\left(\sqrt{7}-5\right)\left(\sqrt{7}+5\right)}-\frac{2\sqrt{7}-10}{\left(\sqrt{7}-5\right)\left(\sqrt{7}+5\right)}=\frac{20}{7-25}=\frac{20}{-18}=\frac{10}{-9}\)
\(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2+\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}=\frac{12+2\sqrt{35}+12-2\sqrt{35}}{2}=\frac{24}{2}=12\)
\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right)\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right)\frac{1}{\sqrt{7}-\sqrt{5}}=\frac{\left(\sqrt{7}+\sqrt{5}\right)}{\sqrt{5}-\sqrt{7}}=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2}{\left(\sqrt{5}-\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)}=\frac{12+2\sqrt{35}}{-2}=-6-\sqrt{35}\)
\(\frac{3}{\sqrt{5}-2}+\frac{2}{\sqrt{5}+3}-\frac{1}{\sqrt{5}+4}=\frac{3\left(\sqrt{5}+2\right)}{5-4}+\frac{2\left(\sqrt{5}-3\right)}{5-9}-\frac{\sqrt{5}-4}{5-16}\)
\(=3\sqrt{5}+6+\frac{2\sqrt{5}-6}{-4}+\frac{4-\sqrt{5}}{-11}=\frac{66\sqrt{5}+132}{22}+\frac{33-11\sqrt{5}}{22}+\frac{2\sqrt{5}-8}{22}\)
\(=\frac{66\sqrt{5}-11\sqrt{5}+2\sqrt{5}+132+33-8}{22}=\frac{57\sqrt{5}+157}{22}\)
\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\frac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\frac{6+2\sqrt{5}}{4}=\frac{32-6-2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}=\frac{14-\sqrt{5}}{2}\) \(\left(\frac{9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2-\left(\frac{9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2=\left(\frac{9-2\sqrt{14}-9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)\left(\frac{9-2\sqrt{14}+9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)=\frac{-72\sqrt{14}}{\sqrt{7}-\sqrt{2}}\)
\(A=\sqrt{11-2\sqrt{10}}+\sqrt{9-2\sqrt{4}}-\sqrt{10}-\sqrt{7}\)
\(=\sqrt{\left(\sqrt{10}-1\right)^2}+\sqrt{5}-\sqrt{10}-\sqrt{7}=\sqrt{10}-1+\sqrt{5}-\sqrt{10}-\sqrt{7}\)
\(=\sqrt{5}-\sqrt{7}-1\)
bài này khó thật , Minh Triều nghĩ ra chưa bạn hiền