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12 tháng 8 2019

Câu 1,2,3 Ez quá rồi :3

Câu 4:

Tổng quát:

\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v

12 tháng 8 2019

Câu 5 ko khác câu 4 lắm :v

Câu 5: 

Tổng quát:

\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v

3 tháng 7 2019

\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(-\sqrt{7}-\sqrt{5}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\frac{\sqrt{5}-\sqrt{7}}{\sqrt{7}+\sqrt{5}}=\frac{\left(\sqrt{5}-\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)}{\left(\sqrt{7}+\sqrt{5}\right)^2}=\frac{2}{12+2\sqrt{35}}\)

3 tháng 7 2019

\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+3\right)}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{8-2\sqrt{15}}{2}+\frac{8+2\sqrt{15}}{2}-\frac{\left(\sqrt{5}+1\right)^2}{4}=8-\frac{6+2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}\)

2 tháng 8 2016

a) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

\(=\left(\sqrt{9\cdot11}-\sqrt{9\cdot2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

\(=\left(3\sqrt{11}-3\sqrt{2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

\(=3\cdot11-3\sqrt{22}-11+3\sqrt{22}\)

\(=33-11=22\)

b)\(3\sqrt{\frac{9}{8}}-\sqrt{\frac{49}{2}}+\sqrt{\frac{25}{18}}\)

\(=\frac{9}{\sqrt{8}}-\frac{7}{\sqrt{2}}+\frac{5}{\sqrt{18}}\)

\(=\frac{9}{2\sqrt{2}}-\frac{7}{\sqrt{2}}+\frac{5}{3\sqrt{2}}\)

\(=\frac{27-42+10}{6\sqrt{2}}\)

\(=-\frac{5}{6\sqrt{2}}\)

c)\(\left(1+\frac{5-\sqrt{5}}{1-\sqrt{5}}\right)\left(\frac{5+\sqrt{5}}{1+\sqrt{5}}+1\right)\)

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

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

\(=1-5=-4\)

 

7 tháng 9 2019

22) \(\frac{1}{\sqrt{5}+\sqrt{2}}+\frac{1}{\sqrt{5}-\sqrt{2}}\)

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

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

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