\(a,=\dfrac{\sqrt{5}+1+\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}}{4}=\dfrac{\sqrt{5}}{2}\\ b,=\sqrt{\left(3-\sqrt{5}\right)^2}+\left|2-\sqrt{5}\right|=3-\sqrt{5}+\sqrt{5}-2=1\\ c,=\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{2}-\dfrac{-\sqrt{3}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}=\sqrt{5}-\sqrt{3}+\sqrt{3}=\sqrt{5}\)

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

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

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

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

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

\(=\dfrac{3-\sqrt{3}}{3}\)

a, \(\dfrac{2}{5}\sqrt{75}-0,5\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\sqrt{12}=2\sqrt{3}-2\sqrt{3}+10\sqrt{3}-\dfrac{4}{3}\sqrt{3}=\dfrac{26}{3}\sqrt{3}\)

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

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

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

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

c, \(3\sqrt{2}-2\sqrt{3}+2\sqrt{3}+3\sqrt{2}=6\sqrt{2}\)

d, \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}+3\right)^2}\)

\(=-\sqrt{6}+3+2\sqrt{6}+3=\sqrt{6}+6\)

e, Ghi đúng đề.

\(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}=\dfrac{a+b-2\sqrt{ab}+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}=2\sqrt{b}\)

a: \(E=1+1=2\)

b: \(=6+3\sqrt{5}+\sqrt{6}-\sqrt{2}+\sqrt{6}-\sqrt{5}\)

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

d: \(=2+\sqrt{3}+2-\sqrt{3}=4\)

1:

\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)

2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)

\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)

\(=\dfrac{20-6}{2}=7\)

a, \(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)

\(=\left|2-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}\right)^2-2\cdot\sqrt{5}\cdot3+3^2}\)

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

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

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

\(=1\)

b, (ĐKXĐ: x ≥ 0; x ≠ 1)

\(A=\dfrac{x-5}{x+2\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{x-5}{x-\sqrt{x}+3\sqrt{x}-3}+\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x-5}{\sqrt{x}\left(\sqrt{x}-1\right)+3\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-1+2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x-5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{3\sqrt{x}+5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

#\(Toru\)

a: \(=\sqrt{5}-2+3-\sqrt{5}=3-2=1\)

b: 

ĐKXĐ: \(x\ge0,x\ne1\)

\(A=\dfrac{x-5+\sqrt{x}-1+2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x+\sqrt{x}-6+2\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

tự làm

a,1/3 nhân căn 6 

b, -9/5 căn 5/2

c, -16/5 căn 15

a) A=căn 6/3

b) B=-1/2 căn 10

c) C=-16/5 căn 15