Question

Các bạn trình bày ra hộ mình nhé

Answer 1

\(G=\left(\dfrac{6-\sqrt{x}}{4x-9}+\dfrac{2}{2\sqrt{x}+3}\right):\dfrac{\sqrt{x}}{2\sqrt{x}-3}=\left(\dfrac{6-\sqrt{x}}{\left(2\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)}+\dfrac{2\left(2\sqrt{x}-3\right)}{\left(2\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)}\right).\dfrac{2\sqrt{x}-3}{\sqrt{x}}=\dfrac{6-\sqrt{x}+4\sqrt{x}-6}{\left(2\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)}.\dfrac{2\sqrt{x}-3}{\sqrt{x}}=\dfrac{3\sqrt{x}}{2\sqrt{x}+3}.\dfrac{1}{\sqrt{x}}=\dfrac{3}{2\sqrt{x}+3}\)Vậy \(G=\dfrac{3}{2\sqrt{x}+3}\)