ĐKXĐ: x≥0; x≠9
Đặt P= \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}\)
P=\(\dfrac{2\sqrt{x}\cdot\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\cdot\left(\sqrt{x}-3\right)}-\dfrac{3-11\sqrt{x}}{x-9}\)
\(P=\dfrac{2x-6\sqrt{x}}{x-9}+\dfrac{x+3\sqrt{x}+\sqrt{x}+3}{x-9}-\dfrac{3-11\sqrt{x}}{x-9}\)
P=\(\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\)
P=\(\dfrac{3x+9\sqrt{x}}{x-9}\)
P=\(\dfrac{3\sqrt{x}\cdot\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-3\right)}\)
P=\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
Vậy P=\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}\) với x≥0: x≠9
\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2x-6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{x+4\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
