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

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

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

a: Sửa đề: \(\dfrac{\sqrt{7-4\sqrt{3}}}{\sqrt{3}-2}\)

\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{3}-2}=\dfrac{2-\sqrt{3}}{\sqrt{3}-2}\)

=-1

b: Sửa đề: \(\dfrac{\sqrt{5-2\sqrt{6}}}{\sqrt{3}-\sqrt{2}}\)

\(=\dfrac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{\sqrt{3}-\sqrt{2}}=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)

=1

giúp mình với ạ 

b: Ta có: \(\dfrac{4}{\sqrt{3}+1}+\dfrac{2}{\sqrt{3}-1}-\dfrac{6}{3-\sqrt{3}}\)

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

\(=2\sqrt{3}-4\)

\(\frac{3^4\cdot5-3^6}{3^4\cdot13+3^4}=\frac{3^4\cdot5-3^2\cdot3^4}{3^4\cdot13+3^4}=\frac{5-3^2}{13}=\frac{5-9}{13}=\frac{-4}{13}\)

Chắc là vậy đó

\(\frac{3^4.5-3^6}{3^4.13+3^4}=\frac{3^4.5-3^4.3^2}{3^4.\left(13+1\right)}=\frac{3^4.\left(5-9\right)}{3^4.14}=-\frac{4}{14}=-\frac{2}{7}\)

a: \(\sqrt[4]{\left(-\dfrac{4}{5}\right)^4}=\left|-\dfrac{4}{5}\right|=\dfrac{4}{5}\)

b: \(\dfrac{\sqrt{4}}{\sqrt{5}}=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}\)

c: \(\left(\sqrt[3]{9}\right)^2=\left(9^{\dfrac{1}{3}}\right)^2=9^{\dfrac{2}{3}}\)

d: \(\sqrt[5]{\sqrt{a}}=\sqrt[5]{a^{\dfrac{1}{2}}}=a^{\dfrac{1}{2}\cdot\dfrac{1}{5}}=a^{\dfrac{1}{10}}\)

e: \(\sqrt[3]{2^6}=\sqrt[3]{\left(2^2\right)^3}=2^2=4\)

`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`

đk : x > = 0; \(3\sqrt{x}-6\ne0\Leftrightarrow x\ne4\)

\(\dfrac{x-4}{3\sqrt{x}-6}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{3\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+2}{3}\)

\(A=29\dfrac{1}{2}\cdot\dfrac{2}{3}+39\dfrac{1}{3}\cdot\dfrac{3}{4}+\dfrac{5}{6}\)

\(=\dfrac{59}{2}\cdot\dfrac{2}{3}+\dfrac{118}{3}\cdot\dfrac{3}{4}+\dfrac{5}{6}\)

\(=\dfrac{59}{3}+\dfrac{118}{4}+\dfrac{5}{6}\)

\(=\dfrac{59}{3}+\dfrac{59}{2}+\dfrac{5}{6}\)

\(=59\cdot\left(\dfrac{1}{3}+\dfrac{1}{2}\right)+\left(\dfrac{1}{3}+\dfrac{1}{2}\right)\)

\(=\dfrac{5}{6}\cdot\left(59+1\right)=\dfrac{5}{6}\cdot60=50\)

Xét: 2^5x7+2^5/2^5.5^2-2^5.3

=2^5x7+2^5x1/2^5x25-2^5x3

=2^5x(7+1)/2^5x(25-3)

=2^5x8/2^5x22

=8/22

=4/11

Xét: 3^4x5-3^6/3^4x13+3^4

=3^4x5-3^4x3^2/3^4x13+3^4

=3^4x(5-9)/3^4x(13+1)

=3^4x(-4)/3^4x14

=-4/14

=-2/7

Mẫu số chung của 4/11 và -2/7 là 77

-->Thừa số phụ tương ứng là 7 và 11

Quy đồng : 4/11=4x7/11x7=28/77

                -2/11=(-2)x11/7x11=-22/77

Vậy ta có 28/77 và -22/77