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9 tháng 7 2021

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

\(=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}=\dfrac{\left(\sqrt{5}+1\right)^2+\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{12}{4}=3\)

b) \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=\dfrac{-\sqrt{2}\left(\sqrt{6}-4\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)

\(=\dfrac{-\sqrt{2}}{\sqrt{3}}-\dfrac{1}{\sqrt{6}}=\dfrac{-3}{\sqrt{6}}=\dfrac{-\sqrt{3}}{\sqrt{2}}\)

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

\(=9+4\sqrt{5}-3=6+4\sqrt{5}\)

d) \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)

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

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

 

9 tháng 7 2021

c) \(\dfrac{2\sqrt{10}+\sqrt{30}-2\sqrt{2}-\sqrt{6}}{2\sqrt{10}-2\sqrt{2}}=1+\dfrac{\sqrt{30}-\sqrt{6}}{2\sqrt{10}-2\sqrt{2}}=1+\dfrac{\sqrt{6}\left(\sqrt{5}-1\right)}{2\sqrt{2}\left(\sqrt{5}-1\right)}\)

\(=1+\dfrac{\sqrt{6}}{2\sqrt{2}}=1+\dfrac{\sqrt{3}}{2}=\dfrac{2+\sqrt{3}}{2}\)

d) \(\sqrt{\left(1-\sqrt{2016}\right)^2}.\sqrt{2017+2\sqrt{2016}}=\left|1-\sqrt{2016}\right|\sqrt{\left(\sqrt{2016}+1\right)^2}\)

\(=\left(\sqrt{2016}+1\right)\left(\sqrt{2016}-1\right)=2015\)

e) \(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}=27+12\sqrt{5}+12\sqrt{5}=27+24\sqrt{5}\)

f) \(2\sqrt{5}\left(2-3\sqrt{5}\right)+\left(1-2\sqrt{5}\right)^2+6\sqrt{5}\)

\(=4\sqrt{5}-30+21-4\sqrt{5}+6\sqrt{5}=6\sqrt{5}-9\)

8 tháng 7 2021

\(2\sqrt{5}\left(2-3\sqrt{5}\right)+\left(1-2\sqrt{5}\right)^2+6\sqrt{5}=4\sqrt{5}-30+20+1-4\sqrt{5}+6\sqrt{5}\)

\(=6\sqrt{5}-9\)

25 tháng 10 2021

1)x>=0

2)v8+5v2-v32+6v1/2=2v2+5v2-4v2+3v2=9v2

3)vx+1=2

x+1=4=>x=3

 

 

6 tháng 12 2021

\(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8\left(\sqrt{5}+1\right)}{4}\\ =2\sqrt{5}-2\sqrt{5}-2=-2\)

Bài 4: 

a: Thay x=36 vào A, ta được:

\(A=\dfrac{6+3}{6-2}=\dfrac{9}{4}\)

b: Ta có: \(B=\dfrac{1}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{2}{x-4}\)

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

\(=\dfrac{x+3\sqrt{x}}{x-4}\)

==>\(\dfrac{3000x+18000}{x\left(x+6\right)}=\dfrac{2650x+5x\left(x+6\right)}{x\left(x+6\right)}\)

=>2650x+5x^2+30x=3000x+18000

=>x=100

6 tháng 6 2021

`a\sqrt(ab) = -\sqrt(a^2 .ab) = -\sqrt(a^3 b)`

`=>` B.

6 tháng 6 2021

\(a\sqrt{ab}=-\sqrt{a^3b}\)

Bài 8:

a: Ta có: \(E=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right):\left(\dfrac{1}{x+1}+\dfrac{x}{x-1}+\dfrac{2}{x^2-1}\right)\)

\(=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-1+x^2+x+2}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{4x}{x^2+2x+1}\)

b: Thay x=3 vào E, ta được:

\(E=\dfrac{4\cdot3}{\left(3+1\right)^2}=\dfrac{12}{4^2}=\dfrac{3}{4}\)

Thay x=-3 vào E, ta được:

\(E=\dfrac{4\cdot\left(-3\right)}{\left(-3+1\right)^2}=\dfrac{-12}{4}=-3\)

9 tháng 10 2021

sai đề rồi ạ