Question

\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}-\dfrac{2x-\sqrt{x}-3}{x-9}\)

mình học yếu các bạn đừng giải tắt nhé 

Answer 1

\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}-\dfrac{2x-\sqrt{x}-3}{x-9}\) ; \(\left(x\ge0;x\ne9\right)\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{2x-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)-\left(2x-\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-\sqrt{x}-3-2x+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)

Answer 2

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)-2x+\sqrt{x}+3}{x-9}\)

\(=\dfrac{x-3\sqrt{x}+2x+5\sqrt{x}-3-2x+\sqrt{x}+3}{x-9}\)

\(=\dfrac{x+3\sqrt{x}}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)