Đặt \(A=\dfrac{15-\sqrt{x}}{\sqrt{x}+2}=\dfrac{\dfrac{15}{2}\left(\sqrt{x}+2\right)-\dfrac{17}{2}\sqrt{x}}{\sqrt{x}+2}=\dfrac{15}{2}-\dfrac{17\sqrt{x}}{2\left(\sqrt{x}+2\right)}\le\dfrac{15}{2}\)
\(A=\dfrac{17-\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=-1+\dfrac{17}{\sqrt{x}+2}>-1\)
\(\Rightarrow-1< A\le\dfrac{15}{2}\Rightarrow A=\left\{0;1;2;3;4;5;6;7\right\}\)
\(\Rightarrow\dfrac{15-\sqrt{x}}{\sqrt{x}+2}=\left\{0;1;2;3;4;5;6;7\right\}\)
\(\Rightarrow x=\left\{225;\dfrac{169}{4};\dfrac{121}{9};\dfrac{81}{16};\dfrac{49}{25};\dfrac{25}{36};\dfrac{9}{49};\dfrac{1}{64}\right\}\)
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