\(B=\left(\dfrac{\left(\sqrt{a}+3\right)^2-\left(\sqrt{a}-3\right)^2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\right)\left(\dfrac{\sqrt{a}-3}{3\sqrt{a}}\right)\)
\(=\left(\dfrac{a+6\sqrt{a}+9-a+6\sqrt{a}-9}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\right)\left(\dfrac{\sqrt{a}-3}{3\sqrt{a}}\right)\)
\(=\dfrac{12\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}.\dfrac{\sqrt{a}-3}{3\sqrt{a}}=\dfrac{4}{\sqrt{a}+3}\)
b.
\(\dfrac{4}{\sqrt{a}+3}\in Z\Rightarrow\sqrt{a}+3=Ư\left(4\right)\)
Mà \(\sqrt{a}+3\ge3\)
\(\Rightarrow\sqrt{a}+3=4\)
\(\Rightarrow a=1\)
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Giúp phần a vs ạ tui mất gốc toán
