`c)root{3}{4}.root{3}{1-sqrt3}.root{6}{(sqrt3+1)^2}`

`=root{3}{4(1-sqrt3)}.root{3}{1+sqrt3}`

`=root{3}{4(1-sqrt3)(1+sqrt3)}`

`=root{3}{4(1-3)}=-2`

`d)2/(root{3}{3}-1)-4/(root{9}-root{3}{3}+1)`

`=(2(root{3}{9}+root{3}{3}+1))/(3-1)-(4(root{3}{3}+1))/(3+1)`

`=root{3}{9}+root{3}{3}+1-root{3}{3}-1`

`=root{3}{9}`

`a)root{3}{8sqrt5-16}.root{3}{8sqrt5+16}`

`=root{3}{(8sqrt5-16)(8sqrt5+16)}`

`=root{3}{320-256}`

`=root{3}{64}=4`

`b)root{3}{7-5sqrt2}-root{6}{8}`

`=root{3}{1-3.sqrt{2}+3.2.1-2sqrt2}-root{6}{(2)^3}`

`=root{3}{(1-sqrt2)^3}-sqrt2`

`=1-sqrt2-sqrt2=1-2sqrt2`

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=1+\sqrt{2}\)

\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{4}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=1+\sqrt{2}\)

Ta có : \(\frac{3}{\sqrt{n}+\sqrt{n+4}}=\frac{3}{4}.\frac{4}{\sqrt{n}+\sqrt{n+4}}=\frac{3}{4}.\frac{4\left(\sqrt{n+4}-\sqrt{n}\right)}{\left(\sqrt{n+4}+\sqrt{n}\right)\left(\sqrt{n+4}-\sqrt{n}\right)}\)

\(=\frac{3}{4}.\frac{4\left(\sqrt{n+4}-\sqrt{n}\right)}{n+4-n}=\frac{3}{4}.\frac{4\left(\sqrt{n+4}-\sqrt{n}\right)}{4}=\frac{3}{4}\left(\sqrt{n+4}-\sqrt{n}\right)\)

Áp dụng ta được :

\(\frac{3}{\sqrt{4}+\sqrt{8}}+\frac{3}{\sqrt{8}+\sqrt{12}}+\frac{3}{\sqrt{12}+\sqrt{16}}+...+\frac{3}{\sqrt{572}+\sqrt{576}}\)

\(=\frac{3}{4}\left(\sqrt{8}-\sqrt{4}+\sqrt{12}-\sqrt{8}+\sqrt{16}-\sqrt{12}+...+\sqrt{576}-\sqrt{572}\right)\)

\(=\frac{3}{4}\left(\sqrt{576}-\sqrt{4}\right)=\frac{3}{4}\left(24-4\right)=\frac{3}{4}.20=15\)

a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=1+\sqrt{2}\)

b: \(\sqrt{\dfrac{2a}{3}}\cdot\sqrt{\dfrac{3a}{8}}=\sqrt{\dfrac{6a^2}{24}}=\sqrt{\dfrac{a^2}{4}}=\dfrac{a}{2}\)

c: \(\sqrt{5a\cdot45a}-3a=-15a-3a=-18a\)

= 2,414213562 nha  Nguyễn Thị Thái Hà ! ! !

k và kb nha hihi ! ! !

Mình cảm ơn bạn 
Mình biết cách làm rồi 

a) Ta có: \(-3\sqrt{16}\cdot\sqrt{90}\)

\(=-3\cdot4\cdot3\sqrt{10}\)

\(=-36\sqrt{10}\)

b) Ta có: \(3\sqrt{\dfrac{4}{3}}-3\sqrt{48}+5\sqrt{75}\)

\(=3\cdot\dfrac{2}{\sqrt{3}}-3\cdot4\sqrt{3}+5\cdot5\sqrt{3}\)

\(=2\sqrt{3}-12\sqrt{3}+25\sqrt{3}\)

\(=15\sqrt{3}\)

c) Ta có: \(4\sqrt[3]{27}-\sqrt[3]{64}-2\sqrt[3]{8}\)

\(=4\cdot3-4-2\cdot2\)

\(=12-4-4=4\)