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8 tháng 4 2021

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

8 tháng 4 2021

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

3 tháng 9 2021

a, \(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

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

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

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

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

b, \(A\in Z\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}\in Z\)

\(\Leftrightarrow\sqrt{x}+3\inƯ_3=\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow\sqrt{x}=0\)

\(\Leftrightarrow x=0\)

3 tháng 9 2021

\(a,A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\left(x\ge0;x\ne9\right)\\ A=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ A=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)

\(b,A\in Z\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}\in Z\Leftrightarrow-3⋮\sqrt{x}+3\\ \Leftrightarrow\sqrt{x}+3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{-6;-4;-2;0\right\}\)

Mà \(\sqrt{x}\ge0\)

\(\Leftrightarrow x\in\left\{0\right\}\)

Vậy \(x=0\) thì A nguyên

 

20 tháng 6 2017

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

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

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

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

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

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

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

21 tháng 6 2021

`a)(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4,x ne 9)`

`=(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`

`=(2sqrtx-9+(sqrtx-3)(sqrtx+3)+(2sqrtx+1)(sqrtx-2))/(x-5sqrtx+6)`

`=(2sqrtx-9+x-9+2x-3sqrtx-2)/(x-5sqrtx+6)`

`=(3x-sqrtx-20)/

21 tháng 6 2021

Lỗi nhẹ :v

16 tháng 9 2018

điều kiện xác định : \(x\ge0;x\ne1\)

a) ta có : \(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\dfrac{1-x}{\sqrt{2}}\right)^2\)

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

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

b) \(x>0\Rightarrow-\sqrt{x}< 0\)\(x< 1\Rightarrow\sqrt{x}-1< 0\)

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

c) ta có : \(P=-\sqrt{x}\left(\sqrt{x}-1\right)=-x+\sqrt{x}=-x+\sqrt{x}-\dfrac{1}{4}+\dfrac{1}{4}\)

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

\(\Rightarrow P_{max}=\dfrac{1}{4}\) khi \(\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\)

vậy GTLN của \(P\)\(\dfrac{1}{4}\) khi \(x=\dfrac{1}{4}\)

21 tháng 10 2021

ĐKXĐ: \(x\ge0,x\ne1\)

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

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

16 tháng 10 2021

ĐKXĐ: \(x\ge0,x\ne1\)

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

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

21 tháng 6 2021

a) P = \(\dfrac{x+2}{\sqrt{x}^3-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\)

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

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

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

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

b) Để \(\dfrac{\sqrt{x}}{x+\sqrt{x}+1}< \dfrac{1}{3}\)

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

<=> \(\dfrac{3\sqrt{x}-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}< 0\)

Mà \(x+\sqrt{x}+1=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

<=> \(-x+2\sqrt{x}-1< 0\)

<=> \(-\left(\sqrt{x}-1\right)^2< 0\) (luôn đúng)

=> P \(< \dfrac{1}{3}\)

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

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

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

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

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

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

13 tháng 6 2018

đkxđ: x≥0; x≠4

\(A=\dfrac{1}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{4-x}\)

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

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

+) A = 1/4 <=> \(\dfrac{2}{2+\sqrt{x}}=\dfrac{1}{4}\Leftrightarrow2+\sqrt{x}=8\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\)(tm)

Vậy x = 36

13 tháng 6 2018

đkxđ \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

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

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

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

để \(A=\dfrac{1}{4}\)

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

\(\Leftrightarrow\sqrt{x}+2=8\)

\(\Leftrightarrow x=36\left(tm\right)\)

vậy tại x=36 thì A=1/4