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11 tháng 7 2018

Bài 1:

a)  \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)

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

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

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

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

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

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

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

c)  ĐK:  \(a\ge0;a\ne1\)

  \(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)

\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)

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

\(=1-a+a=1\)

6 tháng 10 2018

Ai giải giúp mình bài 1 với bài 4 trước đi

18 tháng 8 2016

a/ Với x = \(23-12\sqrt{3}\) ta có:

\(x-11=23-12\sqrt{3}-11=12-12\sqrt{3}=12\left(1-\sqrt{3}\right)\) 

\(\sqrt{x-2}-3=\sqrt{23-12\sqrt{3}-2}-3=\sqrt{21-12\sqrt{3}}-3=\sqrt{3^2-2.3.2\sqrt{3}+\left(2\sqrt{3}\right)^2}-3=\sqrt{\left(3-2\sqrt{3}\right)^2}-3=2\sqrt{3}-6\)                        \(=2\sqrt{3}\left(1-\sqrt{3}\right)\)

=>\(\frac{x-11}{\sqrt{x-2}-3}=\frac{12\left(1-\sqrt{3}\right)}{2\sqrt{3}\left(1-\sqrt{3}\right)}=\frac{12}{2\sqrt{3}}=\frac{2\sqrt{3}.2\sqrt{3}}{2\sqrt{3}}=2\sqrt{3}\)

18 tháng 8 2016

b/ \(\frac{1}{2\left(1+\sqrt{a}\right)}+\frac{1}{2\left(1-\sqrt{a}\right)}-\frac{a^2+2}{1-a^3}=\frac{1-\sqrt{a}}{2\left(1-a\right)}+\frac{1+\sqrt{a}}{2\left(1-a\right)}-\frac{a^2+2}{\left(1-a\right)\left(1-a+a^2\right)}\)

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

Thay : \(a=\sqrt{2}tacó:\)

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

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

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

b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)

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

c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)

\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

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

\(=\dfrac{3}{\sqrt{x}-2}\)