\(p=\frac{\sqrt{x}-2}{\sqrt{x}+1}\) tìm x \(\in\)R để P\(\in\)Z
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LT
9 tháng 8 2019
N=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}+\frac{\sqrt{x}+3}{2-\sqrt{x}}\)
= \(\frac{2\sqrt{x}-9}{x-2\sqrt{x}-3\sqrt{x}+6}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}-\frac{\sqrt{x}+3}{\sqrt{x}-2}\)
= \(\frac{2\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}+\frac{2\sqrt{x}+1}{\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}\)
= \(\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
ĐKXĐ : x ≠ 4 ; x ≠ 9
Rút gọn :
=\(\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
= \(\frac{2\sqrt{x}-9+\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1-\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
=\(\frac{2\sqrt{x}-9+\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
= \(\frac{2\sqrt{x}-9+x-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
= \(\frac{x-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
Để N =5 thì :
<=> \(\frac{x-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\) =5
<=> x-5 = \(\left(5\sqrt{x}-10\right)\left(\sqrt{x}-3\right)\)
<=> x-5 = 5x - \(15\sqrt{x}\) - \(10\sqrt{x}\) +30
<=> x-5x-25\(\sqrt{x}\) =35
TH
9 tháng 8 2019
a) \(\sqrt{x}\ne3;\sqrt{x}\ne2\Rightarrow x\ne4;x\ne9\)
\(N=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}+\frac{\sqrt{x}+3}{2-\sqrt{x}}\)
\(\Leftrightarrow N=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}-\frac{\sqrt{x}+3}{\sqrt{x}-2}\)
\(\Leftrightarrow N=\frac{2\sqrt{x}-9+2x-3\sqrt{x}-2-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(\Rightarrow N=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
b) \(N=5\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-3}=5\)
\(\Leftrightarrow\sqrt{x}+1=5\sqrt{x}-15\Leftrightarrow4\sqrt{x}=16\)
\(\Leftrightarrow\sqrt{x}=4\Rightarrow x=16\) (thỏa mãn)
c) \(N=\frac{\sqrt{x}+1}{\sqrt{x}-5}=\frac{\sqrt{x}-5+6}{\sqrt{x}-5}=1+\frac{6}{\sqrt{x}-5}\)
để N \(\in\) Z thì \(\left(\sqrt{x}-5\right)\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
| \(\sqrt{x}-5\) | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -6 |
| x | 36 | 16 | 49 | 9 | 64 | 4 | 121 | loại |
6 tháng 4 2018
\(P=\left[\frac{1}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(x-1\right)+\left(x-1\right)}\right]\) \(:\frac{\sqrt{x}+1-2}{x-1}\)
\(P=\left[\frac{1}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right]:\frac{\sqrt{x}-1}{x-1}\)
\(P=\left[\frac{1}{\sqrt{x}+1}-\frac{2}{\left(\sqrt{x}+1\right)^2}\right]:\frac{1}{\sqrt{x}+1}\)
\(P=\frac{\sqrt{x}+1-2}{\left(\sqrt{x}+1\right)^2}:\frac{1}{\sqrt{x}+1}\)
\(P=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2}\)
\(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
6 tháng 4 2018
\(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(\Leftrightarrow P=\frac{\sqrt{x}+1-2}{\sqrt{x}+1}\)
\(\Leftrightarrow P=1-\frac{2}{\sqrt{x}+1}\)
để \(P\in Z\) \(\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)\)
\(\Leftrightarrow\sqrt{x}+1\in\left\{\pm1;\pm2\right\}\)
+) \(\sqrt{x}+1=-1\Leftrightarrow\sqrt{x}=-2\) ( vô lí )
+) \(\sqrt{x}+1=1\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)
+) \(\sqrt{x}+1=-2\Leftrightarrow\sqrt{x}=-3\) ( vô lí )
+) \(\sqrt{x}+1=2\Leftrightarrow\sqrt{x}=1\)
vậy để \(P\in Z\) thì \(x\in\left\{1;0\right\}\)
Y
30 tháng 7 2019
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\\x\ne9\end{matrix}\right.\)
\(A=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\\ =\left(\frac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}\right):\left(\frac{9-x+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\\ =\frac{-3}{\sqrt{x}+3}:\frac{4-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\frac{-3}{\sqrt{x}+3}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\frac{3}{\sqrt{x}+2}\)
b) Ta có:
\(P=\frac{3}{\sqrt{x}+2}< 1\\ \Leftrightarrow\frac{3}{\sqrt{x}+2}-1< 0\\ \Leftrightarrow\frac{3-\left(\sqrt{x}+2\right)}{\sqrt{x}+2}< 0\\ \Leftrightarrow\frac{1-\sqrt{x}}{\sqrt{x}+2}< 0\\ \Leftrightarrow1-\sqrt{x}< 0\\ \Leftrightarrow\sqrt{x}>1\\ \Leftrightarrow x>1\)
Vậy với \(x>1;x\ne4;x\ne9\)thì P < 1
c) Để \(A\in Z\Leftrightarrow3⋮\sqrt{x}+2\Leftrightarrow\sqrt{x}+2\inƯ\left(3\right)\)
Ta có bảng sau
| \(\sqrt{x}+2\) | 1 | -1 | 3 | -3 |
| \(\sqrt{x}\) | -1 | -3 | 1 | -5 |
| \(x\) | loại | loại | 1(tm) | loại |
Vậy...................
DT
6 tháng 8 2018
a) e= \(\dfrac{x+\sqrt{x}}{x-2\sqrt{x}+1}:\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{1}{1-\sqrt{x}}+\dfrac{2-x}{x-\sqrt{x}}\right)\)(xkhac0;1)
\(\Leftrightarrow e=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)\(\Leftrightarrow e=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\left(\dfrac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(\Leftrightarrow e=\)\(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow e=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(\Leftrightarrow e=\dfrac{x}{\sqrt{x}-1}\)
vậy e=\(\dfrac{x}{\sqrt{x}-1}\)
b )ta có e>1\(\Leftrightarrow\dfrac{x}{\sqrt{x}-1}>1\)
\(\Leftrightarrow x>\sqrt{x}-1\)
\(\Leftrightarrow x-\sqrt{x}+1>0\)
vì x-\(\sqrt{x}-1>0\) với mọi x khác 0 và khác 1
12 tháng 8 2022
d: Để E là số nguyên thì \(x-1+1⋮\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{1;-1\right\}\)
hay \(x=4\)
e: Để E=9/2 thì \(\dfrac{x}{\sqrt{x}-1}=\dfrac{9}{2}\)
=>\(2x-9\sqrt{x}+9=0\)
=>2x-3 căn x-6 căn x+9=0
=>2 căn x-3=0
hay x=9/4
\(P=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)
Để \(P\) nguyên thì \(\frac{3}{\sqrt{x}+1}\) nguyên
\(\Rightarrow\sqrt{x}+1\in\left\{1;3;-1;-3\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{0;2;-2;-4\right\}\)
\(\Rightarrow x\in\left\{0;\sqrt{2}\right\}\)