Question

Rút gọn biểu thức:\(P=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\)

Answer 1

\(P=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{x-9}\)

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

Answer 2

\(P=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\\ =\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}+\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{3\sqrt{x}}{\sqrt{x}-3}\) 

ĐK : \(x\ge0;x\ne9\)

Answer 3

P= \(\dfrac{2\sqrt{x}.\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right).\left(\sqrt{x}+3\right)+11\sqrt{x}-3}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}\)
  = \(\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}+\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}\)
  = \(\dfrac{3x-9\sqrt{x}}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}\)
  = \(\dfrac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
  = \(\dfrac{3\sqrt{x}}{\sqrt{x}+3}\)