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12 tháng 11 2023

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >1\end{matrix}\right.\)

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

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

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

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

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

b: Để A<0 thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}< 0\)

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

=>\(\sqrt{x}< 1\)

=>0<x<1

12 tháng 11 2023

\(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\left(đkxđ:x>0;x\ne1\right)\\ =\left(\dfrac{\left(\sqrt{x}\right)^3-1^3}{\sqrt{x}\left(\sqrt{x-1}\right)}-\dfrac{\left(\sqrt{x}\right)^3+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\\ =\left(\dfrac{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\\ =\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\\ =\dfrac{4\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\\ =\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)

`b,` Để `A<0` thì :

\(\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-1}< 0\\ \Leftrightarrow\sqrt{x}-1< 0\left(vì.2\left(\sqrt{x}+1\right)>0\right)\\ \Leftrightarrow\sqrt{x}< 1\\ \Leftrightarrow0\le x< 1\)

Kết hợp với điều kiện xác định ta có : \(0< x< 1\)

 

17 tháng 11 2023

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{4;9\right\}\end{matrix}\right.\)

\(A=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

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

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

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

b: A<1

=>A-1<0

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

=>\(\dfrac{\sqrt{x}-5}{\sqrt{x}-2}>0\)

TH1: \(\left\{{}\begin{matrix}\sqrt{x}-5>0\\\sqrt{x}-2>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}>5\\\sqrt{x}>2\end{matrix}\right.\Leftrightarrow\sqrt{x}>5\)

=>x>25

TH2: \(\left\{{}\begin{matrix}\sqrt{x}-5< 0\\\sqrt{x}-2< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}< 2\\\sqrt{x}< 5\end{matrix}\right.\)

=>\(\sqrt{x}< 2\)

=>0<=x<4

10 tháng 10 2021

giúp mk vs

10 tháng 10 2021

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

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

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

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

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

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

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

6 tháng 8 2023

a. ĐK: \(x\ne\pm2\)
\(M=\left[\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x+7}{\left(x-2\right)\left(x+2\right)}\right]:\dfrac{3-x+x-2}{x-2}\)

\(=\dfrac{x^2+2x-\left(x^2-2x+x-2\right)-2x-7}{\left(x-2\right)\left(x+2\right)}.\left(x-2\right)\)

\(=\dfrac{x-5}{x+2}\)

b. \(\dfrac{x-5}{x+2}< 1\Leftrightarrow\dfrac{x-5}{x+2}-1< 0\)

\(\Leftrightarrow\dfrac{-7}{x+2}< 0\Leftrightarrow x+2>0\)

\(\Leftrightarrow x>-2\)
Vậy \(x>-2,x\ne2\)

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)