\(a,\\ ĐK:x\ge0;x\ne4;x\ne9\\ A=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\\ A=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(b,\\ x=\dfrac{4-\sqrt{7}}{2}=\dfrac{8-2\sqrt{7}}{4}=\dfrac{\left(\sqrt{7}-1\right)^2}{4}\\ \to A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\dfrac{\sqrt{7}-1}{2}+1}{\dfrac{\sqrt{7}-1}{2}-3}\\ \to A=\dfrac{\dfrac{\sqrt{7}+1}{2}}{\dfrac{\sqrt{7}-7}{2}}=\dfrac{\sqrt{7}+1}{\sqrt{7}-7}\\ \to A=\dfrac{\left(\sqrt{7}+1\right)\left(\sqrt{7}+7\right)}{42}=\dfrac{7+4\sqrt{7}}{21}\)
\(c,\\ A=\dfrac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\in Z\\ \to\sqrt{x}-3\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\\ \to\sqrt{x}\in\left\{1;2;4;5;7\right\}\left(\sqrt{x}\ge0\right)\\ \to x\in\left\{1;4;16;25;49\right\}\)
\(d,\\ \dfrac{1}{A}=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1-4}{\sqrt{x}+1}=1-\dfrac{4}{\sqrt{x}+1}\\ \to\dfrac{1}{A}\ge1-\dfrac{4}{\sqrt{0}+1}=1-\dfrac{4}{1}=-3\\ "="\Leftrightarrow x=0\)
a: \(A=\dfrac{2\sqrt{x}-9-x+9+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}-2\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
b:Khi x=(4-căn 7)/2=(8-2căn 7)/4 thì
\(A=\left(\dfrac{\sqrt{7}-1}{2}+1\right):\left(\dfrac{\sqrt{7}-1}{2}-3\right)\)
\(=\dfrac{\sqrt{7}+1}{2}:\dfrac{\sqrt{7}-7}{2}=\dfrac{\sqrt{7}+1}{\sqrt{7}-7}=\dfrac{-7-4\sqrt{7}}{21}\)
c: A nguyên
=>căn x-3+4 chia hết cho căn x-3
=>căn x-3 thuộc {1;-1;2;-2;4;-4}
=>x thuộc {16;25;1;49}
