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12 tháng 6 2018
a) \(=\left(\sqrt{3}+2\right)^2\)
b)\(=\left(\sqrt{5}-\sqrt{2}\right)^2\)
c)\(=\left(\sqrt{5}+\sqrt{3}\right)^2\)
d)\(=\left(\sqrt{10}-\sqrt{2}\right)^2\)
e) \(=\left(\sqrt{7}+\sqrt{5}\right)^2\)
21 tháng 8 2020
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: \(B=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}\)
\(=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+\frac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{4}+\sqrt{3}\right)\left(\sqrt{4}-\sqrt{3}\right)}+\frac{\sqrt{5}-\sqrt{4}}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+\frac{\sqrt{6}-\sqrt{5}}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+2-\sqrt{3}+\sqrt{5}-2+\sqrt{6}-\sqrt{5}\)
\(=-1+\sqrt{6}\)
17 tháng 11 2019
\(=\sqrt{\frac{\sqrt{5}\left(8\sqrt{5}-3\sqrt{35}\right)}{\left(8\sqrt{5}+3\sqrt{35}\right)\left(8\sqrt{5}-3\sqrt{35}\right)}}\)\(\left(3\sqrt{2}+\sqrt{14}\right)\)
\(=\sqrt{\frac{40-15\sqrt{7}}{5}}.\left(3\sqrt{2}+\sqrt{14}\right)\)
\(=\sqrt{8-3\sqrt{7}}\left(3\sqrt{2}+\sqrt{14}\right)\)
\(=\frac{\sqrt{2}\sqrt{8-3\sqrt{7}}}{\sqrt{2}}\left(3\sqrt{2}+\sqrt{14}\right)\)
\(=\frac{\sqrt{16-3\sqrt{7}}}{\sqrt{2}}\left(3\sqrt{2}+\sqrt{14}\right)\)
\(=\frac{\sqrt{\left(3-\sqrt{7}\right)^2}}{\sqrt{2}}\left(3\sqrt{2}+\sqrt{14}\right)\)
\(=\frac{\left(3-\sqrt{7}\right)}{\sqrt{2}}.\sqrt{2}\left(3+\sqrt{7}\right)\)
\(=9-7\)
\(=2\)
AT
12 tháng 7 2018
A = \(\sqrt{2}\left(\sqrt{8}-\sqrt{32}-2\sqrt{18}\right)=\sqrt{16}-\sqrt{64}-2\sqrt{36}=4-8-2\cdot6=-4-12=-16\)
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\(B=\sqrt{2}-\sqrt{3-\sqrt{5}}=\dfrac{2-\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\dfrac{2-\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\dfrac{2-\sqrt{5}+1}{\sqrt{2}}=\dfrac{3-\sqrt{5}}{\sqrt{2}}\)
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\(C=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\dfrac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}-\dfrac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\)
còn lại lúc nx mk lm nốt nhé, h bận
9 tháng 7 2022
Bài 1:
a: \(=\sqrt{32.4}=\dfrac{9}{5}\sqrt{10}\)
b: \(=\sqrt{5\cdot5\cdot7\cdot7\cdot11\cdot11}=5\cdot7\cdot11=385\)
c: \(=5-2\sqrt{6}\)
d: \(=18-1=17\)
e: \(=3\sqrt{2}-2\sqrt{3}+7\sqrt{3}-7\sqrt{2}=-4\sqrt{2}+5\sqrt{3}\)