Tìm x ϵ Z để C có giá trị nguyên
\(C=\frac{2\sqrt{x}+3}{\sqrt{x}-1}\)
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6 tháng 11 2021
a)ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x+\sqrt{x}-2\sqrt{x}+2-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
b) \(x=9\Rightarrow A=\dfrac{3}{3+1}=\dfrac{3}{4}\)
\(x=7-4\sqrt{3}\Rightarrow A=\dfrac{\sqrt{7-4\sqrt{3}}}{\sqrt{7-4\sqrt{3}}+1}=\dfrac{\sqrt{7-2\sqrt{12}}}{\sqrt{7-2\sqrt{12}}+1}=\dfrac{\sqrt{4-2\sqrt{3}\sqrt{4}+3}}{\sqrt{4-2\sqrt{3}\sqrt{4}+3}+1}=\dfrac{2-\sqrt{3}}{2-\sqrt{3}+1}=\dfrac{2-\sqrt{3}}{3-\sqrt{3}}=\dfrac{\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}=\dfrac{3-\sqrt{3}}{6}\)
PT
1
25 tháng 10 2016
Để A nguyên thì \(\sqrt{x}-1\inƯ\left(5\right)\)
Mà Ư(5)={1;-1;5;-5}
=> \(\sqrt{x}-1\in\left\{1;-1;5;-5\right\}\)
Ta có bảng sau:
| \(\sqrt{x}-1\) | 1 | -1 | 5 | -5 |
| \(\sqrt{x}\) | 2 | 0 | 6 | -4 |
| x | 4 | 0 | 36 | loại |
Vậy \(x\in\left\{0;4;36\right\}\)
7 tháng 7 2021
a) Ta có: \(A=\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{x-4\sqrt{x}+3}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{2\left(\sqrt{x}-1\right)+2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}:\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{1}{2\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)^2}\)
AH
1
AH
Akai Haruma
Giáo viên
28 tháng 10 2021
Lời giải:
Với $x$ nguyên, để $N$ nguyên thì $\sqrt{x}-5$ là ước của $9$
$\Rightarrow \sqrt{x}-5\in\left\{\pm 1;\pm 3;\pm 9\right\}$
$\Rightarrow \sqrt{x}\in\left\{4; 6; 8; 2; 14; -4\right\}$
Vì $\sqrt{x}\geq 0$ nên: $\sqrt{x}\in\left\{4; 6; 8; 2; 14\right\}$
$\Rightarrow x\in\left\{16; 36; 64; 4; 196\right\}$
6 tháng 7 2021
a) \(Q=\) \(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\left(x>0;x\ne1\right)\)
\(Q=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(Q=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(Q=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(Q=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(Q=\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) \(=\dfrac{2}{x-1}\) \(\left(đpcm\right)\).
b) Để \(Q\in Z\) <=> \(\dfrac{2}{x-1}\in Z\) <=> \(x-1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Ta có bảng sau:
| x -1 | 1 | -1 | 2 | -2 |
| x | 2(TM) | 0(ko TM) | 3(TM) | -1(koTM) |
Vậy để biểu thức Q nhận giá trị nguyên thì \(x\in\left\{2;3\right\}\)
14 tháng 7 2016
a) ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\\x\ne9\end{cases}}\)
b) \(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x-3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}+3}{2\left(\sqrt{x}-1\right)}=\frac{-3\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=-\frac{3}{2\left(\sqrt{x}-3\right)}\)c) Để P nguyên thì \(2\left(\sqrt{x}-3\right)\in\left\{-3;-1;1;3\right\}\)=> x thuộc rỗng.
\(C=\frac{2\sqrt{x}+3}{\sqrt{x}-1}\left(ĐK:x\ge0;x\ne1\right)\\ =\frac{2\left(\sqrt{x}-1\right)+5}{\sqrt{x}-1}=2+\frac{5}{\sqrt{x}-1}\)
Vậy để C nguyên thì \(\sqrt{x}-1\inƯ\left(5\right)\)
Mà Ư(5)={1;-1;5;-5}
=> \(\sqrt{x}-1=\left\{-1;1;5;-5\right\}\)
Ta có bawnngr sau:
Vậy x={0;4;36}
thanks nha