Tìm x ϵ Z để \(B=\dfrac{5}{\sqrt{x}-2}\) nguyên
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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 \(x\in\left\{2;3\right\}\)
24 tháng 12 2023
a: Sửa đề: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne9\end{matrix}\right.\)
Để A là số nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)
=>\(\sqrt{x}-3+4⋮\sqrt{x}-3\)
=>\(4⋮\sqrt{x}-3\)
=>\(\sqrt{x}-3\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(\sqrt{x}\in\left\{4;2;5;1;7;-1\right\}\)
=>\(\sqrt{x}\in\left\{4;2;5;1;7\right\}\)
=>\(x\in\left\{16;4;25;1;49\right\}\)
b: 
17 tháng 2 2021
a) Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\dfrac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right):\left(\dfrac{25-x-\left(x-9\right)+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-\dfrac{\sqrt{x}+5}{\sqrt{x}+5}\right):\left(\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}:\dfrac{x+9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{x+9}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{x+9}\)
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: $\sqrt{x}\in\left\{4; 6; 8; 2; 14\right\}$
$\Rightarrow x\in\left\{16; 36; 64; 4; 196\right\}$
15 tháng 8 2023
a: ĐểA nguyên thì x^2+2x+x+2-3 chia hết cho x+2
=>-3 chia hết cho x+2
=>x+2 thuộc {1;-1;3;-3}
=>x thuộc {-1;-3;1;-5}
b: B nguyên khi x^2+x+3 chia hết cho x+1
=>3 chia hết cho x+1
=>x+1 thuộc {1;-1;3;-3}
=>x thuộc {0;-2;2;-4}
7 tháng 7 2021
a) Ta có: \(A=\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{x-4\sqrt{x}+3}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{2\left(\sqrt{x}-1\right)+2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}:\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{1}{2\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)^2}\)
LT
2
23 tháng 12 2021
\(P=\dfrac{x-5}{x-4}:\dfrac{x-5}{2x}=\dfrac{2x}{x-4}\)
\(\Rightarrow\)\(\dfrac{2x}{x-4}\in Z\)
\(\Rightarrow\)\(\dfrac{2\left(x-4\right)+8}{x-4}\in Z\)
\(\Rightarrow\)\(2+\dfrac{8}{x-4}\in Z\Rightarrow\)\(\dfrac{8}{x-4}\in Z\Rightarrow x-4\inƯ\left(8\right)=\left\{...\right\}\)
Bạn làm tiếp nhé!
1 tháng 7 2021
a,bn viết đúng đề xíu nhé \(\dfrac{\sqrt{a}+2}{\sqrt{a+3}}\) sửa \(\dfrac{\sqrt{a}+2}{\sqrt{a}+3}\)
đk: \(a\ge0,a\ne4\)
=>\(P=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{1}{\sqrt{a}-2}\)
\(=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)\(=\dfrac{\left(\sqrt{a}-4\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b, \(P=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}=1+\dfrac{-2}{\sqrt{a}-2}\) nguyên\(< =>\sqrt{a}-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(=>a\in\left\{9;1;16;0\right\}\)(TM)
5 tháng 7 2021
a) P = \(\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}+\dfrac{1}{2-\sqrt{a}}\left(ĐKXĐ:a\ge0;a\ne4\right)\)
P = \(\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}-\dfrac{1}{\sqrt{a}-2}\)
P = \(\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
P = \(\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
P = \(\dfrac{\left(\sqrt{a}-4\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
P = \(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b) Ta có: P = \(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\) = 1 - \(\dfrac{2}{\sqrt{a}-2}\)
Để \(P\in Z\) <=> 1 - \(\dfrac{2}{\sqrt{a}-2}\) \(\in Z\) <=> \(\sqrt{a}-2\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Ta có bảng sau:
| \(\sqrt{a}-2\) | 1 | -1 | 2 | -2 |
| \(\sqrt{a}\) | 3 | 1 | 4 | 0 |
| a | 9 (TM) | 1 (TM) | 16 (TM) | 0 (TM) |
Vậy để \(P\in Z\) thì \(a\in\left\{0;1;9;16\right\}\)
\(\Leftrightarrow\sqrt{x}-2\in\left\{-1;1;5\right\}\)
hay \(x\in\left\{1;9;49\right\}\)
Lí do