a/ ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)
\(P=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\\ =\left(\frac{1-\sqrt{a^3}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1+\sqrt{a^3}}{1+\sqrt{a}}-\sqrt{a}\right)\\ =\left(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}+a\right)}{1+\sqrt{a}}-\sqrt{a}\right)\\ =\left(1+\sqrt{a}+a+\sqrt{a}\right)\left(1-\sqrt{a}+a-\sqrt{a}\right)\\ =\left(a+2\sqrt{a}+1\right)\left(a-2\sqrt{a}+1\right)\\ =\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)^2\)
\(=\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\\ =\left(a-1\right)^2\\ =a^2-2a+1\)
b/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
\(P=\left(\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\frac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{2\sqrt{x}-2-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\right)\\ =\left(\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\\ =\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}\\ =\frac{-3}{\sqrt{x}+3}\)
Bạn nhớ ktr lại cho chắc nha .-.
