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) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)

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

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

Sao làm hổng ai bảo đú.n/g vậy :(((

Nhân cả tử và mẫu với biểu thức liên hợp của mẫu (câu a mẫu cuối kì kì)

\(A=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\frac{1}{4}=\sqrt{3}-\frac{3}{4}\)

\(B=-\left(\sqrt{1}+\sqrt{2}-\sqrt{2}-\sqrt{3}+\sqrt{3}+\sqrt{4}-...+\sqrt{7}+\sqrt{8}-\sqrt{8}-\sqrt{9}\right)\)

\(B=-\left(\sqrt{1}-\sqrt{9}\right)=2\)

CM : \(\sqrt{\left(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}\right)^2}=1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}\) 

= \(\frac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}=\frac{n^2\left[\left(n+1\right)^2+1\right]+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\) = \(\frac{n^2\left(n^2+2n+2\right)+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\)

=\(\frac{n^4+2n^2\left(n+1\right)+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\) = \(\frac{\left(n^2+n+1\right)^2}{\left(n^2+n\right)^2}\) =>\(\sqrt{\left(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}\right)}=\frac{n^2+n+1}{n^2+n}\)

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

Ta có : 

A = \(\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+\left(1+\frac{1}{4}-\frac{1}{5}\right)+...+\left(1+\frac{1}{2012}-\frac{1}{2013}\right)\)

= 2012 - \(\frac{1}{2013}\) \(\approx\) 2012

\(\frac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)

\(=\frac{\sqrt{35}.(5\sqrt{7}-7\sqrt{5}+2\sqrt{70})}{\sqrt{35}.\sqrt{35}}\)

\(=\frac{\sqrt{35}.(5\sqrt{7}-7\sqrt{5}+2\sqrt{70})}{35}\)

\(\sqrt{\frac{4}{3}}+\sqrt{12}-\frac{4}{3}\sqrt{\frac{3}{4}}\)

\(=\frac{\sqrt{4}}{\sqrt{3}}+\sqrt{12}-\frac{4}{3}\cdot\frac{\sqrt{3}}{\sqrt{4}}\)

\(=\frac{2\sqrt{3}}{\sqrt{3}.\sqrt{3}}+\sqrt{12}-\frac{4}{3}\cdot\frac{\sqrt{3}}{2}\)

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

\(=2\sqrt{3}\left(\frac{1}{3}+1-\frac{1}{3}\right)\)

\(=2\sqrt{3}\)

\(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}+5}\right):2\sqrt{5}\)

\(=\left(5\cdot\frac{\sqrt{1}}{\sqrt{5}}+\frac{1}{2}\sqrt{4.5}-\frac{5}{4}\sqrt{\frac{4+25}{5}}\right)\cdot\frac{1}{2\sqrt{5}}\)

\(=\left(\frac{5\sqrt{5}}{\sqrt{5}.\sqrt{5}}+\frac{1}{2}.2\sqrt{5}-\frac{5}{4}\sqrt{\frac{29}{5}}\right)\cdot\frac{\sqrt{5}}{2\cdot\sqrt{5}\cdot\sqrt{5}}\)

\(=\left(\frac{5\sqrt{5}}{5}+\sqrt{5}-\frac{5}{4}\cdot\frac{\sqrt{29}}{\sqrt{5}}\right)\cdot\frac{\sqrt{5}}{10}\)

\(=\left(\sqrt{5}+\sqrt{5}-\frac{5}{4}\cdot\frac{\sqrt{29}\sqrt{5}}{\sqrt{5}\sqrt{5}}\right)\cdot\frac{\sqrt{5}}{10}\)

\(=\left(2\sqrt{5}-\frac{5}{4}\cdot\frac{\sqrt{145}}{5}\right)\cdot\frac{\sqrt{5}}{10}\)

\(=\left(2\sqrt{5}-\frac{\sqrt{145}}{4}\right)\cdot\frac{\sqrt{5}}{10}\)

a, = \(\frac{\sqrt{7}-5}{2}-\frac{2\left(3-\sqrt{7}\right)}{4}+\frac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\frac{5\left(4-\sqrt{7}\right)}{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}\)

a, = \(=\frac{\sqrt{7}-5}{2}-\frac{3-\sqrt{7}}{2}+\frac{6\sqrt{7}+12}{7-4}-\frac{20-5\sqrt{7}}{16-7}=\frac{\sqrt{7}-5-3+\sqrt{7}}{2}+\frac{6\sqrt{7}+12}{3}-\frac{20-5\sqrt{7}}{9}\)

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