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16 tháng 7 2018

a

6 tháng 8 2017

\(A=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\frac{6\sqrt{x}}{3\sqrt{x}+1}\)

\(A=\left[\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right].\frac{3\sqrt{x}+1}{6\sqrt{x}}\)

\(A=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{6\sqrt{x}}\)

\(A=\frac{3\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}.\frac{1}{6\sqrt{x}}\)

\(A=\frac{\sqrt{x}+1}{6\sqrt{x}-2}\)

\(A=\frac{5}{6}\Leftrightarrow\frac{\sqrt{x}+1}{6\sqrt{x}-2}=\frac{5}{6}\)

\(\Leftrightarrow6\sqrt{x}+6=30\sqrt{x}-10\)

\(\Leftrightarrow24\sqrt{x}=16\)

\(\Leftrightarrow\sqrt{x}=\frac{2}{3}\Leftrightarrow x=\frac{4}{9}\)

20 tháng 8 2018

\(A=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]\div\frac{6\sqrt{x}}{3\sqrt{x}+1}\)

\(A=\left[\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]\times\frac{3\sqrt{x}+1}{6\sqrt{x}}\)

\(A=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}\times\frac{1}{6\sqrt{x}}\)

\(A=\frac{3\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}\times\frac{1}{6\sqrt{x}}\)

\(A=\frac{\sqrt{x}+1}{6\sqrt{x}-2}\)

\(A=\frac{5}{6}\)

\(\Leftrightarrow\frac{\sqrt{x}+1}{6\sqrt{x}-2}=\frac{5}{6}\)

\(\Leftrightarrow6\sqrt{x}+6=30\sqrt{x}-10\)

\(\Leftrightarrow24\sqrt{x}=16\)

\(\Leftrightarrow\sqrt{x}=\frac{2}{3}\)

\(\Leftrightarrow x=\frac{4}{9}\)

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

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

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

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

20 tháng 7 2016

từ dòng cuối là sai rồi bạn à

Bạn bỏ dòng cuối đi còn lại đúng rồi

Ở tử đặt nhân tử chung căn x chung  rồi lại đặt căn x +1 chung

Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra 

rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)

 

21 tháng 7 2016

cảm ơn bạn nha ok

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:  \(a\ge0;a\ne1\)

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

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

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

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

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

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

\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{-x+9+2x-4\sqrt{x}-5}\)

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

23 tháng 5 2021

Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé