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11 tháng 5 2017

1/ ĐKXĐ: \(x\ge0,x\ne1\)

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

= \(\left[\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}-1}\right]-\left(1+\dfrac{\sqrt{x}}{x+1}\right)\)

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

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

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

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

b/ Với \(x\ge0,x\ne1\)

Để \(E=-\dfrac{1}{7}\Leftrightarrow\dfrac{-x-2\sqrt{x}}{x+1}=-\dfrac{1}{7}\)

\(\Leftrightarrow-7x-14\sqrt{x}+x+1=0\)

\(\Leftrightarrow-6x-14\sqrt{x}+1=0\)

\(\Leftrightarrow\left(6\sqrt{x}+7-\sqrt{55}\right)\left(6\sqrt{x}+7+\sqrt{55}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}6\sqrt{x}+7-\sqrt{55}=0\\6\sqrt{x}+7+\sqrt{55}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{-7+\sqrt{55}}{6}\\\sqrt{x}=\dfrac{-7-\sqrt{55}}{6}\left(ktm\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{52-7\sqrt{55}}{18}\)

Vậy để \(E=-\dfrac{1}{7}\) thì \(x=\dfrac{52-7\sqrt{55}}{18}\)

2 tháng 11 2021

\(1,\\ a,E=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ b,E>0\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}>0\Leftrightarrow\sqrt{x}-1>0\left(\sqrt{x}>0\right)\\ \Leftrightarrow x>1\\ 2,\\ a,B=\dfrac{x-\sqrt{x}+\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\left(\sqrt{x}+1\right)\\ B=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,B=2\Leftrightarrow\sqrt{x}-1=2\left(\sqrt{x}+1\right)\\ \Leftrightarrow\sqrt{x}-1=2\sqrt{x}+2\\ \Leftrightarrow\sqrt{x}=-3\Leftrightarrow x\in\varnothing\)

a: Ta có: \(E=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right):\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\)

\(=\left(\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right):\left(\dfrac{x-1}{\sqrt{x}}\right)\)

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

\(=\dfrac{4x^2}{\left(x-1\right)^2}\)

b: Để E=2 thì \(4x^2=2\left(x-1\right)^2\)

\(\Leftrightarrow4x^2-2x^2+4x-2=0\)

\(\Leftrightarrow2x^2+4x-2=0\)

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

\(\Leftrightarrow\left(x+1\right)^2=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{2}-1\\x=\sqrt{2}-1\end{matrix}\right.\)

c: Ta có: \(x=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)^2\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=2\)

Thay x=2 vào E, ta được:

\(E=\dfrac{4\cdot2^2}{1}=16\)

4 tháng 8 2017

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

a) ta có ĐKXĐ của E là x \(\ne\) 1

x \(\ne\) 0

x \(\ne\) -1

b) ( \(\dfrac{\sqrt{x}}{\sqrt{x}-1}\)- \(\dfrac{\sqrt{1}}{x-\sqrt{x}}\)) : ( \(\dfrac{1}{\sqrt{x}+1}\)+\(\dfrac{2}{\sqrt{x}-1}\))

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

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

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

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

= \(\dfrac{1}{\sqrt{x}}\). \(\dfrac{1}{2}\)

= \(\dfrac{1}{2\sqrt{x}}\)

13 tháng 8 2021

a,\(ĐK:x>0,x\ne1,x\ne4\)

\(A=\left[\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\left[\dfrac{x-1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)

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

b,\(x=3-2\sqrt{2}=2-2\sqrt{2}+1=\left(\sqrt{2}-1\right)^2\)

\(=>A=\dfrac{\sqrt{2}-3}{3\sqrt{2}-3}\)

13 tháng 8 2021

a) ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}-1>0\\\sqrt{x}-2>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>1\\x>4\end{matrix}\right.\) \(\Leftrightarrow x>4\)

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

\(=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

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

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

b) Ta có \(x=3-2\sqrt{2}=2-2\sqrt{2}+1=\left(2-1\right)^2=1\)

Thay \(x=1\) vào \(A\), ta được:

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

17 tháng 7 2021

Làm ơn giúp mình với... :(

17 tháng 10 2023

3:

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >9\end{matrix}\right.\)

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

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

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

b: M>1/3

=>M-1/3>0

=>\(\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{3}>0\)

=>\(\dfrac{6-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}>0\)

=>\(3-\sqrt{x}>0\)

=>\(\sqrt{x}< 3\)

=>0<=x<9

c: \(\sqrt{x}+3>=3\) với mọi x thỏa mãn ĐKXĐ

=>\(M=\dfrac{2}{\sqrt{x}+3}< =\dfrac{2}{3}\) với mọi x thỏa mãn ĐKXĐ

Dấu = xảy ra khi x=0

17 tháng 10 2023

bn bt làm câu 2 ko ạ giúp mik với 

5 tháng 9 2023

\(E=\left(\dfrac{x\sqrt{x}}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right)+\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\) (ĐK: \(x\ne1;x>0\))

\(E=\left[\dfrac{\left(\sqrt{x}\right)^3-1^3}{x-\sqrt{x}}-\dfrac{\left(\sqrt{x}\right)^3+1^3}{x+\sqrt{x}}\right]+\left[\dfrac{x}{\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right]\left[\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)

\(E=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]+\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(E=\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\right)+\dfrac{\left(\sqrt{x}\right)^2-1^2}{\sqrt{x}}\cdot\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

\(E=\dfrac{2\sqrt{x}}{\sqrt{x}}+\dfrac{2x+2}{\sqrt{x}}\)

\(E=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)

5 tháng 9 2023

https://hoc24.vn/cau-hoi/rut-gon-cac-bieu-thuc-sau21-a-leftdfracxsqrtx-1x-1-dfracx-1sqrtx-1right-leftsqrtx-dfracsqrtxsqrtx-1right-22-a-leftdfracxsqrtx.8357393071878

ĐKXĐ: \(x\ge0;x\ne1\)

Ta có: \(A=\left(2+\dfrac{2x+\sqrt{x}}{2\sqrt{x}+1}\right)\left(2-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)

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

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

26 tháng 4 2022

\(=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\left(dk:x\ne0,\pm1\right)\)

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

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

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