Question

A=\(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{3}{x\sqrt{x}+1}+\dfrac{2}{x-\sqrt{x}+1}\)

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

Answer 1

Bạn ơi đề bài là j

Answer 2

a: \(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\dfrac{-3+2\sqrt{x}+2}{x\sqrt{x}+1}\)

\(=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\dfrac{2\sqrt{x}-1}{x\sqrt{x}+1}\)

\(=\dfrac{2x^2+2\sqrt{x}-9x\sqrt{x}-9+2x\sqrt{x}-10x+12\sqrt{x}-x+5\sqrt{x}-6}{\left(x\sqrt{x}+1\right)\left(x-5\sqrt{x}+6\right)}\)

\(=\dfrac{2x^2+19\sqrt{x}-7x\sqrt{x}-11x-15}{\left(x\sqrt{x}+1\right)\left(x-5\sqrt{x}+6\right)}\)

b: \(=\dfrac{x-\sqrt{x}+1-3+2\sqrt{x}+2}{x\sqrt{x}+1}\)

\(=\dfrac{x+\sqrt{x}}{x\sqrt{x}+1}=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)