Question

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

a) Rút gọn A

b) So sánh A với 1,5

Answer 1

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

\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}{2x+3\sqrt{x}+1}\)

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

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

\(=\dfrac{6\sqrt{x}-10-6\sqrt{x}-3}{2\left(2\sqrt{x}+1\right)}=\dfrac{-13}{2\left(2\sqrt{x}+1\right)}< 0\)

=>A<3/2