Question

Answer 1

Sửa đề: \(\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\)

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

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

\(=\dfrac{x\sqrt{x}-3-2x+6\sqrt{x}+6\sqrt{x}-18-x-4\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x\left(\sqrt{x}-3\right)+8\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}=\dfrac{x+8}{\sqrt{x}+1}\)