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4 tháng 9 2020

Bài 1 : 

a, \(\frac{2ab}{\sqrt{a}+\sqrt{b}}=\frac{2ab\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\frac{2ab\left(\sqrt{a}-\sqrt{b}\right)}{a-b}\)

b, \(\frac{2\sqrt{10}-5}{4-\sqrt{10}}=\frac{\left(2\sqrt{10}-5\right)\left(4+\sqrt{10}\right)}{\left(4-\sqrt{10}\right)\left(4+\sqrt{10}\right)}=\frac{\sqrt{10}}{2}\)

c, \(\frac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}=\frac{\left(9-2\sqrt{3}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}=\frac{\sqrt{6}}{2}\)

4 tháng 9 2020

2. \(\frac{1}{\sqrt{3}-\sqrt{2}}-\frac{2}{\sqrt{7}+\sqrt{5}}-\frac{3}{\sqrt{7-2\sqrt{10}}}+\frac{4}{\sqrt{10+2\sqrt{21}}}\)

\(=\frac{\sqrt{3}+\sqrt{2}}{3-2}-\frac{2.\left(\sqrt{7}-\sqrt{5}\right)}{7-5}-\frac{3}{\sqrt{2-2\sqrt{10}+5}}+\frac{4}{\sqrt{3+2\sqrt{21}+7}}\)

\(=\left(\sqrt{3}+\sqrt{2}\right)-\frac{2\left(\sqrt{7}-\sqrt{5}\right)}{2}-\frac{3}{\sqrt{\left(\sqrt{2}-\sqrt{5}\right)^2}}+\frac{4}{\sqrt{\left(\sqrt{3}+\sqrt{7}\right)^2}}\)

\(=\left(\sqrt{3}+\sqrt{2}\right)-\left(\sqrt{7}-\sqrt{5}\right)-\frac{3}{\left|\sqrt{2}-\sqrt{5}\right|}+\frac{4}{\left|\sqrt{3}+\sqrt{7}\right|}\)

\(=\left(\sqrt{3}+\sqrt{2}\right)-\left(\sqrt{7}-\sqrt{5}\right)-\frac{3}{\sqrt{5}-\sqrt{2}}+\frac{4}{\sqrt{3}+\sqrt{7}}\)

\(=\left(\sqrt{3}+\sqrt{2}\right)-\left(\sqrt{7}-\sqrt{5}\right)-\frac{3.\left(\sqrt{5}+\sqrt{2}\right)}{5-2}+\frac{4.\left(\sqrt{7}-\sqrt{3}\right)}{7-3}\)

\(=\left(\sqrt{3}+\sqrt{2}\right)-\left(\sqrt{7}-\sqrt{5}\right)-\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}+\frac{4\left(\sqrt{7}-\sqrt{3}\right)}{4}\)

\(=\left(\sqrt{3}+\sqrt{2}\right)-\left(\sqrt{7}-\sqrt{5}\right)-\left(\sqrt{5}+\sqrt{2}\right)+\left(\sqrt{7}-\sqrt{3}\right)\)

\(=\sqrt{3}+\sqrt{2}-\sqrt{7}+\sqrt{5}-\sqrt{5}-\sqrt{2}+\sqrt{7}-\sqrt{3}=0\)

2 tháng 7 2019

\(a,\frac{2\sqrt{10}-5}{4-\sqrt{10}}=\frac{\left(2\sqrt{10}-5\right)\left(4+\sqrt{10}\right)}{\left(4-\sqrt{10}\right)\left(4+\sqrt{10}\right)}=\frac{20+6\sqrt{10}-5\sqrt{10}-9}{16-10}.\)

\(=\frac{11-\sqrt{10}}{6}\)

\(b,=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{54-8}\)

\(=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{46}\)

27 tháng 9 2016

a. \(\frac{26}{5-2\sqrt{3}}\)=\(\frac{26\cdot\left(5+2\sqrt{3}\right)}{\left(5-2\sqrt{3}\right)\left(5+2\sqrt{3}\right)}\)=\(\frac{26\cdot\left(5+2\sqrt{3}\right)}{5^2-\left(2\sqrt{3}\right)^2}=\frac{26\cdot\left(5+2\sqrt{3}\right)}{13}=2\cdot\left(5+2\sqrt{3}\right)=10+4\sqrt{3}\)

b.\(\frac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}=\frac{\sqrt{3}\cdot\left(3\sqrt{3}-2\right)}{\sqrt{2}\cdot\left(3\sqrt{3}-2\right)}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{6}}{2}\)

c.\(\frac{2\sqrt{10}-5}{4-\sqrt{10}}=\frac{\sqrt{5}\cdot\left(2\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}\cdot\left(2\sqrt{2}-\sqrt{5}\right)}=\frac{\sqrt{5}}{\sqrt{2}}=\frac{\sqrt{10}}{2}\)

d.\(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)=\(4\sqrt{5}\)

 

3 tháng 8 2020

a/ \(\frac{1}{2+\sqrt{3}}-\frac{1}{2-\sqrt{3}}+5\sqrt{3}\)

\(=\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}-\frac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+5\sqrt{3}\)

\(=\frac{2-\sqrt{3}}{4-3}-\frac{2+\sqrt{3}}{4-3}+5\sqrt{3}\)

\(=2-\sqrt{3}-2-\sqrt{3}+5\sqrt{3}\)

\(=3\sqrt{3}\)

Vậy..

3 tháng 8 2020

b/ \(\frac{1}{\sqrt{5}+2}-\sqrt{9+4\sqrt{5}}\)

\(=\frac{1}{\sqrt{5}+2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)

\(=\frac{1}{\sqrt{5}+2}-\left|\sqrt{5}+2\right|\)

\(=\frac{\sqrt{5}-2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\sqrt{5}-2\)

\(=\sqrt{5}-2-\sqrt{5}-2\)

\(=-4\)

Vậy..

7 tháng 8 2019
https://i.imgur.com/B4LOqxT.jpg
12 tháng 8 2019

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