Bạn đăng 1 lần thôi nhé!
\(a,A=\dfrac{9+2}{9+3}=\dfrac{11}{12}\\ b,P=AB=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\cdot\dfrac{x+4\sqrt{x}+4-x+4\sqrt{x}-4+x-2\sqrt{x}+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{1}{\sqrt{x}+3}\cdot\dfrac{x+6\sqrt{x}+9}{\sqrt{x}-2}=\dfrac{\left(x+3\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\\ c,P=\dfrac{\sqrt{x}-2+5}{\sqrt{x}-2}=1+\dfrac{5}{\sqrt{x}-2}\in Z\\ \Leftrightarrow\sqrt{x}-2\inƯ\left(5\right)=\left\{-1;1;5\right\}\left(\sqrt{x}-2\ge-2\right)\\ \Leftrightarrow\sqrt{x}\in\left\{1;3;7\right\}\\ \Leftrightarrow x\in\left\{1;9;49\right\}\)
