Question

\(A=\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\) với \(x\ge0\), \(x\ne9\)

a) Rút gọn A.

b) Tính giá trị của A với \(x=14-6\sqrt{5}\)

Answer 1

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

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

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

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

\(A=\dfrac{-2\sqrt{x}-12}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)