\(1,\\ a,A=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-1\right]:\dfrac{4-x+x-4-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\\ A=\left[\dfrac{\sqrt{x}}{\sqrt{x}+2}-1\right]\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{9-x}\\ A=\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}+2}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\)
\(b,A>-\dfrac{1}{2}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}+\dfrac{1}{2}>0\Leftrightarrow\dfrac{4+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}>0\\ \Leftrightarrow\dfrac{\sqrt{x}+7}{2\left(\sqrt{x}+3\right)}>0\left(luôn.đúng\right)\)
Vậy \(A>-\dfrac{1}{2},\forall x\)
\(c,A=\dfrac{2}{\sqrt{x}+3}\in Z\\ \Leftrightarrow\sqrt{x}+3\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}\in\varnothing\left(\sqrt{x}+3\ge3\right)\)
Vậy ko có gt nào của x nguyên để A nguyên
\(2,\\ a,A=\dfrac{\sqrt{\left(3+2\sqrt{2}\right)^2}-5\sqrt{\left(3-2\sqrt{2}\right)^2}}{\sqrt{2}-1}\\ A=\dfrac{3+2\sqrt{2}-15+10\sqrt{2}}{\sqrt{2}-1}=\dfrac{12\sqrt{2}-12}{\sqrt{2}-1}\\ A=\dfrac{12\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=12\\ b,\sin^2\alpha+\cos^2\alpha=1\Leftrightarrow\sin^2\alpha=1-\dfrac{1}{9}=\dfrac{8}{9}\\ \Leftrightarrow\sin\alpha=\dfrac{2\sqrt{2}}{3}\\ \Leftrightarrow B=\dfrac{\dfrac{2\sqrt{2}}{3}-1}{\dfrac{2\sqrt{2}}{3}+\dfrac{2}{3}}=\dfrac{2\sqrt{2}-3}{3}\cdot\dfrac{2\sqrt{2}+2}{3}\\ \Leftrightarrow B=\dfrac{8+4\sqrt{2}-6\sqrt{2}-6}{3}=\dfrac{2-2\sqrt{2}}{3}\)
