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\(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\left(x\ge0;x\ne9\right)=\dfrac{\sqrt{x}+3-2}{\sqrt{x}+3}=1-\dfrac{2}{\sqrt{x}+3}\)
Để \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\in Z\Leftrightarrow\dfrac{2}{\sqrt{x}+3}\in Z\)
\(\Leftrightarrow2⋮\sqrt{x}+3\\ \Leftrightarrow\sqrt{x}+3\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{-5;-4;-2;-1\right\}\\ \Leftrightarrow x\in\left\{1;4;16;25\right\}\)
Vậy \(x\in\left\{1;4;16;25\right\}\) thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\in Z\)
Tick plz
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}+3\ne0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne-3\left(loại\right)\end{matrix}\right.\)\(\Rightarrow x\ge0\)
\(x\in Z\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\in Z\Rightarrow\left(\sqrt{x}+1\right)⋮\left(\sqrt{x}+3\right)\)
\(\Rightarrow\left(\sqrt{x}+3-2\right)⋮\left(\sqrt{x}+3\right)\)
Vì \(\Rightarrow\left(\sqrt{x}+3\right)⋮\left(\sqrt{x}+3\right)\)
\(\Rightarrow2⋮\left(\sqrt{x}+3\right)\Rightarrow\sqrt{x}+3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Ta có bảng:
| \(\sqrt{x}+3\) | -1 | -2 | 1 | 2 |
| \(x\) | \(\sqrt{x}=-4\left(loại\right)\) | \(\sqrt{x}=-5\left(loại\right)\) | \(\sqrt{x}=-2\left(loại\right)\) | \(\sqrt{x}=-1\left(loại\right)\) |
Vậy không có x nguyên thỏa mãn đề bài
a: \(P=\left(\dfrac{2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Để P nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{-1;1;2\right\}\)
hay \(x\in\left\{0;4;9\right\}\)
\(P=A\cdot B\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\cdot\dfrac{2\sqrt{x}+6+x-3\sqrt{x}+3-5\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}-3\right)}\cdot\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}-3\right)^2}=\dfrac{2\sqrt{x}}{\sqrt{x}+3}\)
Để P nguyên thì
\(2\sqrt{x}⋮\sqrt{x}+3\)
\(\Leftrightarrow2\sqrt{x}+6-6⋮\sqrt{x}+3\)
=>\(\sqrt{x}+3\inƯ\left(-6\right)\)
=>\(\sqrt{x}+3\in\left\{3;6\right\}\)
=>\(\sqrt{x}\in\left\{0;3\right\}\)
=>\(x\in\left\{0;9\right\}\)
Kết hợp ĐKXĐ, ta được: x=0
Để M là số nguyên thì \(12\sqrt{x}+5⋮3\sqrt{x}-1\)
=>\(12\sqrt{x}-4+9⋮3\sqrt{x}-1\)
=>\(3\sqrt{x}-1\in\left\{1;-1;3;-3;9;-9\right\}\)
=>\(3\sqrt{x}\in\left\{2;0;4;10\right\}\)
=>\(\sqrt{x}\in\left\{0;\dfrac{2}{3};\dfrac{4}{3};\dfrac{10}{3}\right\}\)
mà x là số chính phương
nên x=0
\(M=\dfrac{12\sqrt{x}+5}{3\sqrt{x}-1}\)
\(M=\dfrac{12\sqrt{x}-4+9}{3\sqrt{x}-1}\)
\(M=\dfrac{4\left(3\sqrt{x}-1\right)+9}{3\sqrt{x}-1}\)
\(M=\dfrac{4\left(3\sqrt{x}-1\right)}{3\sqrt{x}-1}+\dfrac{9}{3\sqrt{x}-1}\)
\(M=4+\dfrac{9}{3\sqrt{x}-1}\)
M nguyên khi:
\(9\) ⋮ \(3\sqrt{x}-1\)
Mà: \(3\sqrt{x}-1\ge-1\)
\(\Rightarrow3\sqrt{x}-1\in\left\{1;-1;3;9\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{\dfrac{2}{3};0;\dfrac{4}{3};\dfrac{10}{3}\right\}\)
\(\Rightarrow x\in\left\{\dfrac{4}{9};0;\dfrac{16}{9};\dfrac{100}{9}\right\}\)
Mà: x là số chính phương nên:
x = 0
ĐKXĐ: x>=0
Để A là số nguyên thì \(\sqrt{x}+13⋮\sqrt{x}+5\)
=>\(\sqrt{x}+5+8⋮\sqrt{x}+5\)
=>\(\sqrt{x}+5\inƯ\left(8\right)\)
mà \(\sqrt{x}+5>=5\)
nên \(\sqrt{x}+5=8\)
=>x=9
ĐK: \(x\ge0\)
Để \(\dfrac{\sqrt{x}+13}{\sqrt{x}+5}\) có giá trị nguyên
Mà: \(\dfrac{\sqrt{x}+13}{\sqrt{x}+5}=\dfrac{\sqrt{x}+5+8}{\sqrt{x}+5}\)
\(=\dfrac{\sqrt{x}+5}{\sqrt{x}+5}+\dfrac{8}{\sqrt{x}+5}=1+\dfrac{8}{\sqrt{x}+5}\)
Vậy: \(8\) ⋮ \(\sqrt{x}+5\)
\(\Rightarrow\sqrt{x}+5\inƯ\left(8\right)=\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
Mà: \(\sqrt{x}+5\ge5\)
\(\Rightarrow\sqrt{x}+5\in\left\{8\right\}\)
\(\Rightarrow x=9\left(tm\right)\)
ĐKXĐ: \(x>0;x\ne9\)
\(P=\left(\dfrac{x+7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}-\dfrac{4\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\right)\left(\dfrac{\sqrt{x}+6}{\sqrt{x}}\right)\)
\(=\left(\dfrac{x+7-4\sqrt{x}-4+\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\right)\left(\dfrac{\sqrt{x}+6}{\sqrt{x}}\right)\)
\(=\left(\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\right).\left(\dfrac{\sqrt{x}+6}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}.\dfrac{\left(\sqrt{x}+6\right)}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+6}{\sqrt{x}+1}\)
b.
Ta có \(P=\dfrac{\sqrt{x}+1+5}{\sqrt{x}+1}=1+\dfrac{5}{\sqrt{x}+1}\)
Do \(\sqrt{x}+1>0\Rightarrow\dfrac{5}{\sqrt{x}+1}>0\Rightarrow P>1\)
\(P=\dfrac{6\left(\sqrt{x}+1\right)-5\sqrt{x}}{\sqrt{x}+1}=6-\dfrac{5\sqrt{x}}{\sqrt{x}+1}\)
Do \(\left\{{}\begin{matrix}5\sqrt{x}>0\\\sqrt{x}+1>0\end{matrix}\right.\) ;\(\forall x>0\Rightarrow\dfrac{5\sqrt{x}}{\sqrt{x}+1}>0\)
\(\Rightarrow P< 6\Rightarrow1< P< 6\)
Mà P nguyên \(\Rightarrow P=\left\{2;3;4;5\right\}\)
- Để \(P=2\Rightarrow\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=2\Rightarrow\sqrt{x}+6=2\sqrt{x}+2\Rightarrow x=16\)
- Để \(P=3\Rightarrow\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=3\Rightarrow\sqrt{x}+6=3\sqrt{x}+3\Rightarrow\sqrt{x}=\dfrac{3}{2}\Rightarrow x=\dfrac{9}{4}\)
- Để \(P=4\Rightarrow\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=4\Rightarrow\sqrt{x}+6=4\sqrt{x}+4\Rightarrow\sqrt{x}=\dfrac{2}{3}\Rightarrow x=\dfrac{4}{9}\)
- Để \(P=5\Rightarrow\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=5\Rightarrow\sqrt{x}+6=5\sqrt{x}+5\Rightarrow\sqrt{x}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{16}\)
\(P\in Z\Rightarrow3P\in Z\Rightarrow\dfrac{3\sqrt{x}+15}{3\sqrt{x}+1}\in Z\)
\(\Rightarrow1+\dfrac{14}{3\sqrt{x}+1}\in Z\)
\(\Rightarrow3\sqrt{x}+1=Ư\left(14\right)=\left\{1;2;7;14\right\}\) (do \(3\sqrt{x}+1\ge1\))
\(3\sqrt{x}+1=1\Rightarrow x=0\)
\(3\sqrt{x}+1=2\Rightarrow x=\dfrac{1}{9}\notin Z\) (loại)
\(3\sqrt{x}+1=7\Rightarrow x=4\)
\(3\sqrt{x}+1=14\Rightarrow x=\dfrac{169}{9}\notin Z\) (loại)
Thế \(x=\left\{0;4\right\}\) vào P đều thỏa mãn
Vậy ....