\((\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\...">
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30 tháng 8 2020

a) Ta có:

\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right)\div\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)

\(P=\frac{x-1}{\left(\sqrt{x}-1\right)\sqrt{x}}\div\frac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(P=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\sqrt{x}}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)

\(P=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}=\frac{x-1}{\sqrt{x}}\)

30 tháng 8 2020

b) Ta có: \(P>0\)

\(\Leftrightarrow\frac{x-1}{\sqrt{x}}>0\)

\(\Leftrightarrow\frac{\left(x-1\right)\sqrt{x}}{x}>0\)

\(\Rightarrow\left(x-1\right)\sqrt{x}>0\)

\(\Rightarrow\hept{\begin{cases}x-1>0\\\sqrt{x}>0\end{cases}}\Rightarrow x>1\)

Vậy khi \(x>1\Leftrightarrow P>0\)

c) Ta có: \(P=6\)

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

\(\Leftrightarrow x-1=6\sqrt{x}\)

\(\Leftrightarrow\left(x-1\right)^2=36x\)

\(\Leftrightarrow x^2-38x+1=0\)

\(\Leftrightarrow\left(x^2-38x+361\right)-360=0\)

\(\Leftrightarrow\left(x-19\right)^2-\left(6\sqrt{10}\right)^2=0\)

\(\Leftrightarrow\left(x-19-6\sqrt{10}\right)\left(x-19+6\sqrt{10}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-19-6\sqrt{10}=0\\x-19+6\sqrt{10}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=19+6\sqrt{10}\\x=19-6\sqrt{10}\end{cases}}\)

7 tháng 7 2017

a, ĐK \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

\(P=\frac{x-1}{\sqrt{x}}:\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

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

Ta thấy \(P=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}>0\forall x>0,x\ne1\)

b, P=\(\frac{x+2\sqrt{x}+1}{\sqrt{x}-1}=\frac{\frac{2}{2+\sqrt{3}}+2\sqrt{\frac{2}{2+\sqrt{3}}}+1}{\sqrt{\frac{2}{2+\sqrt{3}}}-1}\)

=\(\frac{\frac{4}{\left(\sqrt{3}+1\right)^2}+2.\sqrt{\left(\frac{2}{\left(\sqrt{3}+1\right)^2}\right)}+1}{\sqrt{\left(\frac{2}{2+\sqrt{3}}\right)^2}-1}=\frac{\frac{4}{\left(\sqrt{3}+1\right)^2}+2.\frac{2}{\sqrt{3}+1}+1}{\frac{2}{\sqrt{3}+1}-1}\)

\(=\frac{12+6\sqrt{3}}{1-3}=-6-3\sqrt{3}\)

7 tháng 7 2017

cậu ơi câu c đâu ạ??

13 tháng 9 2020

Bài 1.

\(B=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\div\frac{x}{x-\sqrt{x}}\)với \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

a) \(B=\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\div\frac{x}{x-\sqrt{x}}\)

\(B=\left(\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\div\frac{x}{x-\sqrt{x}}\)

\(B=\left(\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\div\frac{x}{x-\sqrt{x}}\)

\(B=\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\div\frac{x}{x-\sqrt{x}}\)

\(B=\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{x}\)

\(B=\frac{4\sqrt{x}\cdot\sqrt{x}}{\left(\sqrt{x}+1\right)x}=\frac{4x}{\left(\sqrt{x}+1\right)x}=\frac{4}{\sqrt{x}+1}\)

b) Để B > 1

=> \(\frac{4}{\sqrt{x}+1}>0\)( với \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\))

Vì 4 > 0

=> \(\sqrt{x}+1>0\)

<=> \(\sqrt{x}>-1\)( luôn luôn đúng \(\forall\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)) ( theo ĐKXĐ )

Vậy \(\forall\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)thì B > 1

Chưa chắc lắm ... Còn câu 2 thì tí nữa mình làm cho 

13 tháng 9 2020

Bài 2.

\(A=2\sqrt{5}-1\)

\(B=\frac{2}{x-1}\cdot\sqrt{\frac{x^2-2x+1}{4x^2}}\)( x > 0 )

a) \(B=\frac{2}{x-1}\cdot\frac{\sqrt{x^2-2x+1}}{\sqrt{4x^2}}\)

\(B=\frac{2}{x-1}\cdot\frac{\sqrt{\left(x-1\right)^2}}{\sqrt{\left(2x\right)^2}}\)

\(B=\frac{2}{x-1}\cdot\frac{\left|x-1\right|}{\left|2x\right|}\)

\(B=\frac{2}{x-1}\cdot\frac{x-1}{2x}=\frac{1}{x}\)( vì x > 0 )

b) Để A + B = 0

=> \(\left(2\sqrt{5}-1\right)+\frac{1}{x}=0\)( ĐKXĐ : \(x\ne0\))

<=> \(\frac{1}{x}=-\left(2\sqrt{5}-1\right)\)

<=> \(\frac{1}{x}=1-2\sqrt{5}\)

<=> \(x\times\left(1-2\sqrt{5}\right)=1\)

<=> \(x=\frac{1}{1-2\sqrt{5}}\)( tmđk )

Vậy \(x=\frac{1}{1-2\sqrt{5}}\)

Bài 1 : 

a )\(A=\frac{3-\sqrt{3}}{\sqrt{3}-1}+\frac{\sqrt{35}-\sqrt{15}}{\sqrt{5}}-\sqrt{28}\)

\(A=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\frac{\sqrt{5}\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{5}}-\sqrt{28}\)

\(A=\sqrt{3}+\sqrt{7}-\sqrt{3}-\sqrt{28}\)

\(A=\sqrt{7}-\sqrt{28}\)

\(A=\sqrt{7}-2\sqrt{7}=-\sqrt{7}\)

Vậy \(A=-\sqrt{7}\)

b)\(B=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{\sqrt{a}+\sqrt{b}}{a-b}\left(a,b>0;a\ne b\right)\)

\(B=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\frac{\sqrt{a}+\sqrt{b}}{a-b}\)

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

\(B=a-b\)

Vậy \(B=a-b\left(a,b>0;a\ne b\right)\)

_Minh ngụy_

Bài 2 :

a )\(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\left(x>0\right)\)

\(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\frac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\frac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

Vậy \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\left(x>0\right)\)

b) \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\left(x>0\right)\)

Ta có : \(B>0\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}>0\)

Vì : \(\sqrt{x}\ge0\forall x\Rightarrow\)để \(B>O\)cần \(\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)( thỏa mãn \(x>0\))

Vậy \(x>1\)thì \(B>0\)

_Minh ngụy_

23 tháng 7 2019

\(a,\)\(T=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)

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

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

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

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

18 tháng 7 2018

a)  \(A=\left(\sqrt{6}+\sqrt{10}\right).\left(\sqrt{5}-\sqrt{3}\right)\)

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

         \(=2\sqrt{2}\)

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

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

       \(=\frac{4}{x-4}+1\)

       \(=\frac{4}{x-4}+\frac{x-4}{x-4}=\frac{x}{x-4}\)