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20 tháng 7 2017
a) điều kiện \(x>0;x\ne1\)
\(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{1-\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(\Leftrightarrow\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(\Leftrightarrow\left(\dfrac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\left(\sqrt{x}+1\right)^2+\left(1-\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(\Leftrightarrow\dfrac{x+2\sqrt{x}+1+x-\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x+2\sqrt{x}+1+\sqrt{x}-1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow\dfrac{x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow\dfrac{x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{4\sqrt{x}}=\dfrac{x+1}{4\sqrt{x}}\)
28 tháng 10 2022
a: \(P=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
b: \(\dfrac{1}{P}-3=\dfrac{x+\sqrt{x}+1-3\sqrt{x}}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)
=>1/P>3
23 tháng 10 2018
điều kiện \(x\ge0;x\ne1\)
a) ta có : \(P=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\)
\(\Leftrightarrow P=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\) \(\Leftrightarrow P=\dfrac{\left(x+2\right)\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)\left(x-1\right)-\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}{\left(x-1\right)\left(x+\sqrt{x}+1\right)}\) \(\Leftrightarrow P=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}\right)}{\left(x-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(x-1\right)}{\left(x-1\right)\left(x+\sqrt{x}+1\right)}\)\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
b) ta có : \(\dfrac{1}{P}-3=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-3=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\) \(\left(x\ne1\right)\)
vậy \(\dfrac{1}{P}-3>0\Leftrightarrow\dfrac{1}{P}>3\)
2 tháng 9 2022
Bài 2:
a: \(A=\dfrac{2x+6\sqrt{x}-x-9\sqrt{x}}{x-9}=\dfrac{x-3\sqrt{x}}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+5\right)}{x-25}=\dfrac{\sqrt{x}}{\sqrt{x}-5}\)
b: \(P=A:B=\dfrac{\sqrt{x}}{\sqrt{x}+3}:\dfrac{\sqrt{x}}{\sqrt{x}-5}=\dfrac{\sqrt{x}-5}{\sqrt{x}+3}\)
\(P-1=\dfrac{\sqrt{x}-5-\sqrt{x}-3}{\sqrt{x}+3}=\dfrac{-8}{\sqrt{x}+3}< 0\)
=>P<1
AN
18 tháng 12 2017
a) P = \(\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}\) - \(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) + \(\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)
ĐK : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
P = \(\dfrac{3x+3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\) - \(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) - \(\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
P = \(\dfrac{3x+3\sqrt{x}-3-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3x+3\sqrt{x}-3-\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\) = \(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\) = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b) Để \(\sqrt{P}\) có nghĩa P ≥ 0 ⇒ \(\sqrt{x}-1\) > 0 ⇒ x = 1
P = \(1+\dfrac{2}{\sqrt{x}-1}>1\)
Xét \(P-\sqrt{P}\) = \(\sqrt{P}\left(\sqrt{P}-1\right)\)
Mà \(\sqrt{P}>0\)
Vì P > 1 ⇒ \(\sqrt{P}>\sqrt{1}\Rightarrow\sqrt{P}>1\Rightarrow\sqrt{P}-1>0\Rightarrow P-\sqrt{P}>0\Leftrightarrow P>\sqrt{P}\)
c) Tìm x để \(\dfrac{1}{P}\in Z\)
\(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\Rightarrow\dfrac{1}{P}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\)
\(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+1\ge1\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\le\dfrac{2}{1}\Leftrightarrow-\dfrac{2}{\sqrt{x}+1}\ge-2\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}\ge-1\)
\(\Rightarrow-1\le\dfrac{1}{P}< 1\Rightarrow\dfrac{1}{P}\in\left\{-1;0\right\}\)
\(với\dfrac{1}{P}=-1\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=-1\)
\(\Leftrightarrow\sqrt{x}-1=-\sqrt{x}-1\)
\(\Leftrightarrow2\sqrt{x}=0\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\)
\(với\dfrac{1}{P}=0\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=0\)
\(\Leftrightarrow\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\left(loại\right)\)
Vậy x=0 thì \(\dfrac{1}{P}\in Z\)
CHÚC BẠN HỌC TỐT 
20 tháng 9 2020
ĐK: \(x\ge0;x\ne1\)
a) \(P=\frac{\sqrt{x}+x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(P=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
\(P=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}.\frac{1}{\sqrt{x}+1}\)
\(P=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
Để \(P=\sqrt{x}\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-1}=\sqrt{x}\Leftrightarrow\sqrt{x}+1=\sqrt{x}\left(\sqrt{x}-1\right)\)\(\sqrt{x}+1\Leftrightarrow x-\sqrt{x}\Leftrightarrow-x+2\sqrt{x}+1=0\)
\(\Leftrightarrow-\left(x-2\sqrt{x}+1\right)+2=0\Leftrightarrow\left(\sqrt{x}-1\right)^2=2\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-1=\sqrt{2}\\\sqrt{x}-1=-\sqrt{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=\sqrt{2}+1\\\sqrt{x}=-\sqrt{2}+1\end{cases}\Leftrightarrow}x=3\pm2\sqrt{2}}\)
b) Với \(x>1\)thì \(P>0\)
Ta dễ thấy \(P=\frac{\sqrt{x}+1}{\sqrt{x}-1}>1\)
Ta có: \(P>0;P>1\)\(\Rightarrow P\left(P-1\right)>0\Leftrightarrow P^2>P\Leftrightarrow P>\sqrt{P}\)
16 tháng 11 2022
a: \(A=\dfrac{\sqrt{x}-1+\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(\dfrac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}+\dfrac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{1+x\sqrt{x}}\right)\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(2x+\sqrt{x}-1\right)\cdot\left(\dfrac{1}{1-x}+\dfrac{\sqrt{x}}{1+x\sqrt{x}}\right)\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left[\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\cdot\dfrac{1+x\sqrt{x}+\sqrt{x}-x\sqrt{x}}{\left(1-x\right)\left(1+x\sqrt{x}\right)}\right]\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left[\dfrac{\left(2\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}{\left(1-\sqrt{x}\right)\left(1+x\sqrt{x}\right)}\right]\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}\cdot\dfrac{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b: Khi x=17-12 căn 2 thì \(A=\dfrac{17-12\sqrt{2}+3-2\sqrt{2}+1}{3-2\sqrt{2}}=7\)
3 tháng 12 2022
a: \(P=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
c: Để \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\) là số nguyên thì \(\sqrt{x}+1-2⋮\sqrt{x}+1\)
=>\(\sqrt{x}+1\in\left\{1;2\right\}\)
=>x=0
AT
25 tháng 7 2018
dkxd: x≥0; x≠1
a/ \(P=1:\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(=1:\left[\dfrac{1}{\sqrt{x}^3-1}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{x+\sqrt{x}+1}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(=1:\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=1:\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=1:\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=1:\dfrac{\sqrt{x}}{x+\sqrt{x}+1}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
TN
25 tháng 7 2018
c. Để \(\sqrt{P}\) có nghĩa \(P\ge0\Rightarrow\sqrt{x}>0\Rightarrow x\ge1\)
Vì \(x\ge1\) nên \(P=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}>1\)
Ta có: P>1 \(P-\sqrt{P}=\sqrt{P}\left(\sqrt{P}-1\right)\left(1\right)\)
Vì P > 1\(\Rightarrow\sqrt{P}>1\Rightarrow\sqrt{P}-1>0\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\sqrt{P}\left(\sqrt{P}-1\right)>0\Rightarrow P-\sqrt{P}>0\Rightarrow P>\sqrt{P}\)