P=\(\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{1-\sqrt{x}}\right):\left(\dfrac{3}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{x-1}\right)\)
a) Rút gọn P
b) Tìm x để P = \(\sqrt{x}-1\)
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23 tháng 9 2021
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{9;4\right\}\end{matrix}\right.\)
b: Ta có: \(P=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
\(=\dfrac{2\sqrt{x}-9+2x-3\sqrt{x}-2-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
28 tháng 6 2023
a: \(P=\dfrac{x+2+x-1}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{2x+1-x+\sqrt{x}-1}{x\sqrt{x}+1}=\dfrac{x+\sqrt{x}}{x\sqrt{x}+1}=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)
b: |P|=2/3
mà P>=0
nên P=2/3
=>căn x/(x-căn x+1)=2/3
=>2x-2căn x+2=3căn x
=>2x-5*căn x+2=0
=>(căn x-2)(2căn x-1)=0
=>x=4 hoặc x=1/4
19 tháng 1 2022
a, x > 0 ; x khác 1
\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{x-\sqrt{x}}\right):\dfrac{1}{\sqrt{x}-1}\)
\(=\left(\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{1}{\sqrt{x}-1}=\dfrac{x-2}{\sqrt{x}}\)
b, Ta có : \(P=\dfrac{x-2}{\sqrt{x}}=1\Rightarrow x-2=\sqrt{x}\)
\(\Leftrightarrow x-\sqrt{x}-2=0\Leftrightarrow\left(\sqrt{x}+1>0\right)\left(\sqrt{x}-2\right)=0\Leftrightarrow x=4\)(tm)
19 tháng 1 2022
a: \(P=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{1}=\dfrac{x-2}{\sqrt{x}}\)
b: Để P=1 thì \(x-\sqrt{x}-2=0\)
hay x=4
Y
4 tháng 4 2022
\(a,\)
\(=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3}{3\sqrt{x}+1}\right)\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}+3x}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)
Vậy \(P=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)
\(b,\)Thay \(P=\dfrac{6}{5}\) vào pt, ta có :
\(\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}=\dfrac{6}{5}\)
\(\Leftrightarrow5\left(3\sqrt{x}+1\right)=6\left(3\sqrt{x}-1\right)\)
\(\Leftrightarrow15\sqrt{x}+5-18\sqrt{x}+6=0\)
\(\Leftrightarrow-3\sqrt{x}+11=0\)
\(\Leftrightarrow-3\sqrt{x}=-11\)
\(\Leftrightarrow\sqrt{x}=\dfrac{11}{3}\)
\(\Leftrightarrow x=\left(\dfrac{11}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{121}{9}\)
Vậy \(x=\dfrac{121}{9}\) thì \(P=\dfrac{6}{5}\)
17 tháng 1 2022
ĐKXĐ: \(x\ge0,x\ne1\)
a) \(P=\left[\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}-\dfrac{1}{\sqrt{x}-1}\right]:\left(\dfrac{x+1+\sqrt{x}}{x+1}\right)\)
\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}.\dfrac{x+1}{x+\sqrt{x}+1}=\dfrac{-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{1-\sqrt{x}}{x+\sqrt{x}+1}\)
\(P=\dfrac{1-\sqrt{x}}{x+\sqrt{x}+1}< 0\Leftrightarrow1-\sqrt{x}< 0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)
31 tháng 8 2023
a: \(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+1}\cdot\dfrac{2}{\sqrt{x}+3}=-\dfrac{6}{\sqrt{x}+3}\)
b: P>=-1/2
=>P+1/2>=0
=>\(\dfrac{-6}{\sqrt{x}+3}+\dfrac{1}{2}>=0\)
=>\(\dfrac{-12+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}>=0\)
=>căn x-9>=0
=>x>=81
c: căn x+3>=3
=>6/căn x+3<=6/3=2
=>-6/căn x+3>=-2
Dấu = xảy ra khi x=0
26 tháng 8 2021
đk : \(x\ge0,x\ne1\)
\(=>P=\left[\dfrac{2\left(\sqrt{x}+2\right)-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right]:\left[\dfrac{x+\sqrt{x}-2+3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right]\)
\(P=\left[\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right].\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+1}\right]\)
\(P=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b,\(x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\) thay vào P
\(=>P=\dfrac{2\sqrt{\left(\sqrt{5}-1\right)^2}-1}{\sqrt{\left(\sqrt{5}-1\right)^2}+1}=\dfrac{2\sqrt{5}-3}{\sqrt{5}}\)
c,\(=>\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}=>2x-\sqrt{x}=\sqrt{x}+1\)
\(=>2x-2\sqrt{x}-1=0< =>2\left(x-\sqrt{x}-\dfrac{1}{2}\right)=0\)
\(=>x-\sqrt{x}-\dfrac{1}{2}=>\Delta=1-4\left(-\dfrac{1}{2}\right)=3>0=>\left[{}\begin{matrix}x1=\dfrac{1+\sqrt{3}}{2}\\x2=\dfrac{1-\sqrt{3}}{2}\end{matrix}\right.\)
đối chiếu đk loại x2 còn x1 thỏa
12 tháng 1 2022
\(a,P=\dfrac{-x+2\sqrt{x}-1+x-2\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}:\dfrac{2\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ P=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}+1}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\\ \Rightarrow P=\dfrac{\sqrt{5}-1}{\sqrt{5}-1+1}=\dfrac{\sqrt{5}-1}{\sqrt{5}}=\dfrac{5-\sqrt{5}}{5}\\ c,\dfrac{P}{\sqrt{x}}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\cdot\dfrac{1}{\sqrt{x}}=\dfrac{1}{\sqrt{x}-1}\le\dfrac{1}{0-1}=-1\)
Vậy \(\left(\dfrac{P}{\sqrt{x}}\right)_{max}=-1\Leftrightarrow x=0\)
a: \(P=\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}:\dfrac{3\sqrt{x}+3+\sqrt{x}-1-2}{x-1}\)
\(=\dfrac{-4\sqrt{x}}{3\sqrt{x}}=\dfrac{-4}{3}\)
b: Để \(P=\sqrt{x}-1\) thì \(\sqrt{x}=-\dfrac{1}{3}\)(vô lý)
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