Question

A=(\(\dfrac{2\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}-1}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\))(\(3\sqrt{x}-\dfrac{\sqrt{x}+4}{\sqrt{x}-1}\))

a,rút gọn A                                        b,tìm x để A<2

Answer 1

a: \(A=\dfrac{2x-6\sqrt{x}+\sqrt{x}-3-2x+4\sqrt{x}+\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{3x-3\sqrt{x}-\sqrt{x}-4}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-1}{3x-4\sqrt{x}-4}\)

\(=\dfrac{1}{\sqrt{x}-2}\cdot\dfrac{3x-6\sqrt{x}+2\sqrt{x}-4}{\sqrt{x}-1}=\dfrac{3\sqrt{x}+2}{\sqrt{x}-1}\)

b: Để A<2 thì \(\dfrac{3\sqrt{x}+2-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)}< 0\)

=>x<1

Answer 2

=>x<1