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a) Ta có: \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=\left(-\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=-2+2\sqrt{5}-\sqrt{5}\)
\(=-2+\sqrt{5}\)
b) \(\left(\frac{1}{2}\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right)\div\frac{1}{8}\)
\(=\left(\frac{\sqrt{2}}{4}-\frac{3\sqrt{2}}{2}+8\sqrt{2}\right)\cdot8\)
\(=\frac{27\sqrt{2}}{4}\cdot8\)
\(=54\sqrt{2}\)
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) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)
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tu lam di cau nao kho thi hoi hoi vay ko ai tra loi cho dau
cau e)
\(A=\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}\)(suy ra A>=0)
\(A^2=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)\)
\(A^2=1\)
A=1
(bai toan co nhieu cach)
cau m)
\(=\frac{\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
\(=\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{2}}\)
\(=1\)
cau G)
\(=\frac{5\sqrt{7}}{\sqrt{35}}-\frac{7\sqrt{5}}{\sqrt{35}}+\frac{2\sqrt{70}}{\sqrt{35}}\)
\(=\frac{5}{\sqrt{5}}-\frac{7}{\sqrt{7}}+2\sqrt{2}\)
\(=\sqrt{5}-\sqrt{7}+2\sqrt{2}\)
a) Ta có: \(A=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\left(\sqrt{9}-\sqrt{4}\right)\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)
\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)(Vì \(\sqrt{5}>\sqrt{3}\))
\(=5-3-\sqrt{5}\)
\(=2-\sqrt{5}\)
b) Ta có: \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)
\(=\left(\frac{\sqrt{3}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}+\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}+\sqrt{6}\right)\)
\(=\sqrt{3}+\sqrt{3}+\sqrt{6}\)
\(=2\sqrt{3}+\sqrt{6}\)
c) Ta có: \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)
\(=2\sqrt{3}+\sqrt{4-2\cdot2\cdot\sqrt{3}+3}+\sqrt{\frac{1}{3}:3}-\sqrt{\frac{4}{3}:3}+\sqrt{3:3}\)
\(=2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\frac{1}{9}}-\sqrt{\frac{4}{9}}+\sqrt{1}\)
\(=2\sqrt{3}+\left|2-\sqrt{3}\right|+\frac{1}{3}-\frac{2}{3}+1\)
\(=2\sqrt{3}+2-\sqrt{3}+\frac{2}{3}\)(Vì \(2>\sqrt{3}\))
\(=\sqrt{3}+\frac{8}{3}\)
d) Ta có: \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)
\(=\left(\frac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\right)\cdot\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\)
\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\cdot\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\frac{60}{20}\cdot\left|2-\sqrt{3}\right|\)
\(=3\cdot\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))
\(=6-3\sqrt{3}\)
a) Ta có: \(A=\frac{8+2\sqrt{15}+\sqrt{21}+\sqrt{35}}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\frac{\left(\sqrt{3}+\sqrt{5}\right)^2+\sqrt{7}\cdot\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\frac{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}+\sqrt{7}\right)}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\sqrt{3}+\sqrt{5}\)
b) Ta có: \(B=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}\)
\(=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+\frac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{4}+\sqrt{3}\right)\left(\sqrt{4}-\sqrt{3}\right)}+\frac{\sqrt{5}-\sqrt{4}}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+\frac{\sqrt{6}-\sqrt{5}}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+2-\sqrt{3}+\sqrt{5}-2+\sqrt{6}-\sqrt{5}\)
\(=-1+\sqrt{6}\)