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Đặt \(A=\dfrac{15-\sqrt{x}}{\sqrt{x}+2}=\dfrac{\dfrac{15}{2}\left(\sqrt{x}+2\right)-\dfrac{17}{2}\sqrt{x}}{\sqrt{x}+2}=\dfrac{15}{2}-\dfrac{17\sqrt{x}}{2\left(\sqrt{x}+2\right)}\le\dfrac{15}{2}\)
\(A=\dfrac{17-\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=-1+\dfrac{17}{\sqrt{x}+2}>-1\)
\(\Rightarrow-1< A\le\dfrac{15}{2}\Rightarrow A=\left\{0;1;2;3;4;5;6;7\right\}\)
\(\Rightarrow\dfrac{15-\sqrt{x}}{\sqrt{x}+2}=\left\{0;1;2;3;4;5;6;7\right\}\)
\(\Rightarrow x=\left\{225;\dfrac{169}{4};\dfrac{121}{9};\dfrac{81}{16};\dfrac{49}{25};\dfrac{25}{36};\dfrac{9}{49};\dfrac{1}{64}\right\}\)
Để biểu thức đề bài cho có giá trị nguyên thì \(5\sqrt{x}-6⋮2\sqrt{x}-3\)
\(\Leftrightarrow10\sqrt{x}-12⋮2\sqrt{x}-3\)
\(\Leftrightarrow2\sqrt{x}-3\in\left\{-3;-1;1;3\right\}\)
\(\Leftrightarrow2\sqrt{x}\in\left\{0;2;4;6\right\}\)
hay \(x\in\left\{0;1;4;9\right\}\)
\(a,A=\dfrac{2\sqrt{x}-2+2\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\left(x\ge0;x\ne1;x\ne9\right)\\ A=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
\(b,A\in Z\Leftrightarrow\dfrac{\sqrt{x}-3+5}{\sqrt{x}-3}\in Z\Leftrightarrow1+\dfrac{5}{\sqrt{x}-3}\in Z\\ \Leftrightarrow\sqrt{x}-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ Mà.x\ge0\\ \Leftrightarrow\sqrt{x}\in\left\{2;4;8\right\}\\ \Leftrightarrow x\in\left\{4;16;64\right\}\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\\x\ne1\end{matrix}\right.\)
\(A=\dfrac{2\sqrt{x}-2+2\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
b) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}=1+\dfrac{5}{\sqrt{x}-3}\in Z\)
\(\Rightarrow\sqrt{x}-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
Kết hợp đk
\(\Rightarrow x\in\left\{4;16;64\right\}\)
Để A nguyên thì \(2\sqrt{x}+3⋮3\sqrt{x}-1\)
\(\Leftrightarrow6\sqrt{x}+9⋮3\sqrt{x}-1\)
\(\Leftrightarrow3\sqrt{x}-1\in\left\{-1;1;11\right\}\)
\(\Leftrightarrow3\sqrt{x}\in\left\{0;12\right\}\)
hay \(x\in\left\{0;16\right\}\)
\(\dfrac{\sqrt{x}-2}{\sqrt{x}+8}\left(x\ge0\right)=\dfrac{\sqrt{x}+8-10}{\sqrt{x}+8}=1-\dfrac{10}{\sqrt{x}+8}\)
Để biểu thức nguyên thì \(10⋮\sqrt{x}+8\)
\(\Leftrightarrow\sqrt{x}+8\inƯ\left(10\right)=\left\{-10;-5;-2;-1;1;2;5;10\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{-18;-13;-10;-9;-7;-6;-3;2\right\}\)
Mà \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}\in\left\{2\right\}\Leftrightarrow x=4\)
Với \(x\ge0\) có : \(\dfrac{\sqrt{x}-2}{\sqrt{x}+8}=\dfrac{\sqrt{x}+8-10}{\sqrt{x}+8}=1-\dfrac{10}{\sqrt{x}+8}\)
Để bthuc nhận gt nguyên thì :
\(\dfrac{10}{\sqrt{x}+8}\in Z\Leftrightarrow10⋮\sqrt{x}+8\) \(\Leftrightarrow\sqrt{x}+8\inƯ\left(10\right)\Leftrightarrow\sqrt{x}+8=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Mà \(\sqrt{x}\ge0\) nên ta chỉ xét \(\sqrt{x}+8=\left\{1;2;5;10\right\}\)
Ta có bảng sau
Vậy x=4 là gtri cần tìm