Question

rút gọn B

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

Answer 1

\(ĐKXĐ\) \(x\) khác 9

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

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

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

\(B=\dfrac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{x-9}\)

\(B=\dfrac{-6\sqrt{x}-6}{x-9}\)

 

Answer 2

ĐK: x ≥ 0, x ≠ 9

Ta có: \(B=\dfrac{3\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\left(2x+3\right)}{9-x}\)

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

\(=\dfrac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-6\sqrt{x}-6}{x-9}\)

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