`~rai~`

\(33)\dfrac{a+b+2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}.\quad(1)\\ĐKXĐ:\begin{cases}a\ge 0\\b\ge 0\\\sqrt{a}+\sqrt{b}\ne 0\end{cases}\\\Leftrightarrow \begin{cases}a\ge 0\\b\ge 0\\\sqrt{a}\ne 0\\\sqrt{b}\ne 0\end{cases}\\\Leftrightarrow \begin{cases}a\ge 0\\b\ge 0\\a\ne 0\\b\ne 0\end{cases}\\\Leftrightarrow \begin{cases}a>0\\b>0.\end{cases}\\\text{Khi đó,ta có:}\\(1)=\dfrac{a+b+2\sqrt{ab}-a+b}{\sqrt{a}+\sqrt{b}}\\\quad=\dfrac{2b+2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\\\quad=\dfrac{2\sqrt{b}(\sqrt{b}+\sqrt{a})}{(\sqrt{a}+\sqrt{b})}\\\quad=2\sqrt{b}.\\34)\dfrac{(2+\sqrt{a})^2-(\sqrt{a}-3)^2}{2a-\sqrt{a}}.\quad (1)\\ĐKXĐ:\begin{cases}a\ge0\\2a-\sqrt{a}\ne 0\end{cases}\\\Leftrightarrow \begin{cases}a\ge 0\\\sqrt{a}(2\sqrt{a}-1)\ne0\end{cases}\\\Leftrightarrow \begin{cases}a\ge0\\\sqrt{a}\ne 0\\2\sqrt{a}-1\ne 0\end{cases}\\\Leftrightarrow \begin{cases}a\ge 0\\a\ne 0\\\sqrt{a}\ne\dfrac{1}{2}\end{cases}\\\Leftrightarrow \begin{cases}a>0\\a\ne\dfrac{1}{4}.\end{cases}\\\text{Khi đó,ta có:}\\(1)=\dfrac{(4+4\sqrt{a}+a)-(a-6\sqrt{a}+9)}{2a-\sqrt{a}}\\\quad=\dfrac{4+4\sqrt{a}+a-a+6\sqrt{a}-9}{2a-\sqrt{a}}\\\quad=\dfrac{10\sqrt{a}-5}{2a-\sqrt{a}}\\\quad=\dfrac{5(2\sqrt{a}-1)}{\sqrt{a}(2\sqrt{a}-1)}\\\quad=\dfrac{5}{\sqrt{a}}\\\quad=\dfrac{5\sqrt{a}}{a}.\)

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