Question

Answer 1

\(a,A=\dfrac{17}{3+2}=\dfrac{17}{5}\\ b,B=\dfrac{x+2\sqrt{x}+\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x+3\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ B=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\\ c,P=A:B=\dfrac{17}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+5}=\dfrac{17}{\sqrt{x}+5}\in Z\\ \Leftrightarrow\sqrt{x}+5\inƯ\left(17\right)=\left\{17\right\}\left(\sqrt{x}+5\ge5\right)\\ \Leftrightarrow\sqrt{x}=11\Leftrightarrow x=121\left(tm\right)\)