mik chưa chắc nó chính xác đâu nên bn thông cảm cho mik nha
ĐKXĐ: \(x>0;x\ne9\)
\(D=\left(\dfrac{3}{\sqrt{a}-3}-\dfrac{\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}+1\right):\dfrac{\sqrt{a}}{\sqrt{a}+3}\)
\(=\left(\dfrac{3\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}-\dfrac{\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}+1\right):\dfrac{\sqrt{a}}{\sqrt{a}+3}\)
\(=\left(\dfrac{3\sqrt{a}+9-\sqrt{a}+a-9}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\right):\dfrac{\sqrt{a}}{\sqrt{a}+3}\)
\(=\left(\dfrac{a+2\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\right):\dfrac{\sqrt{a}}{\sqrt{a}+3}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}.\dfrac{\left(\sqrt{a}+3\right)}{\sqrt{a}}\)
\(=\dfrac{\sqrt{a}+2}{\sqrt{a}-3}\)
