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

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

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

b: Thay \(x=3+2\sqrt{2}\) vào P, ta được:

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

a: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{\sqrt{x}-1+1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{5}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}}=\dfrac{x-1}{\sqrt{x}}+\dfrac{5}{\sqrt{x}}=\dfrac{x+4}{\sqrt{x}}\)

b: Để A=5 thì \(x+4=5\sqrt{x}\)

=>x=1(loại) hoặc x=16(nhận)

6 tháng 8 2021

a, A= \(\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\frac{\left(\sqrt{x}\right)^2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{x}{\sqrt{x}+2}\right)\)

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

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

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

b, A >= \(\frac{1}{3\sqrt{x}}\)

=> \(\frac{1}{x+2\sqrt{x}}\) >= \(\frac{1}{3\sqrt{x}}\)

=> x <= -1 , x >= 4 (x khác 0)

a: Ta có: \(A=\left(1+\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}-x}\right)+\dfrac{5}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{\sqrt{x}-1+1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{5}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-1}{1}+\dfrac{5}{\sqrt{x}}\)

\(=\dfrac{x+4}{\sqrt{x}}\)

b: Để A=5 thì \(x+4=5\sqrt{x}\)

\(\Leftrightarrow x=16\)

31 tháng 8 2021

a. \(A=\left(1+\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}-x}\right)+\dfrac{5}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{1-\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}+\dfrac{5}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}.\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{-\sqrt{x}}+\dfrac{5}{\sqrt{x}}\)

\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}+\dfrac{5}{\sqrt{x}}=\dfrac{x-1+5}{\sqrt{x}}=\dfrac{x+4}{\sqrt{x}}\)

b. \(A=5\Leftrightarrow\dfrac{x+4}{\sqrt{x}}=5\Leftrightarrow x+4=5\sqrt{x}\Leftrightarrow x-5\sqrt{x}+4=0\)

\(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=1\end{matrix}\right.\)

Vậy tất cả các x thỏa ycbt là x=1 hoặc x=16

c. \(A>4\Leftrightarrow\dfrac{x+4}{\sqrt{x}}>4\Leftrightarrow\dfrac{x+4}{\sqrt{x}}-4>0\Leftrightarrow\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}}>0\)

Vì \(\left(\sqrt{x}-2\right)^2\ge0\forall x\) nên \(\left\{{}\begin{matrix}\sqrt{x}-2\ne0\\\sqrt{x}>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne4\\x>0\end{matrix}\right.\)

Vậy tất cả các x thỏa mãn ycbt là x>0 và \(x\ne4\)

 

12 tháng 8 2021

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

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

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

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

b) \(\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{1}{3}\)

\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}\)

\(\Leftrightarrow3\sqrt{x}-\sqrt{x}=3\)

\(\Leftrightarrow2\sqrt{x}=3\)

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

\(\Rightarrow x=\dfrac{9}{4}\)

1 tháng 7 2023

\(a,A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\left(dk:x\ge0,x\ne1\right)\)

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

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

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

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

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

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

\(b,x-3\sqrt{x}+2=0\Leftrightarrow x-\sqrt{x}-2\sqrt{x}+2=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{x}-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=4\left(tm\right)\end{matrix}\right.\)

Thay \(x=4\) vào A :

\(A=\dfrac{\sqrt{4}-1}{\sqrt{4}+1}=\dfrac{2-1}{2+1}=\dfrac{1}{3}\)