Câu 1:
\(P=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{2-\sqrt{a}}=\sqrt{a}+2+\sqrt{a}+2=2\sqrt{a}+4\\ A=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ A=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\\ C=\dfrac{\sqrt{x}-1+\sqrt{x}+1-4\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{-2\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ C=\dfrac{-2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2}{1-\sqrt{x}}\)
\(D=\dfrac{\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{2}=\dfrac{2\sqrt{x}}{2}=\sqrt{x}\\ P=\dfrac{8\sqrt{x}-4x+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{\sqrt{x}-2-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ P=\dfrac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-2}=\dfrac{4x}{\sqrt{x}-2}\\ Q=\dfrac{\left(\sqrt{a}+4\right)^2}{\sqrt{a}+4}+\dfrac{\left(3-\sqrt{a}\right)\left(3+\sqrt{a}\right)}{3-\sqrt{a}}-\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{\sqrt{a}}\\ Q=\sqrt{a}+4+3+\sqrt{a}-\sqrt{a}+2\\ Q=\sqrt{a}+9\)
