a,[tex]A= \frac{\sqrt{x}(\sqrt{x-2})}{(\sqrt{x})^3-1}+\frac{\sqrt{x}+1}{\sqrt{x}(x+\sqrt{x}+1)}+\frac{1+2\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}[(\sqrt{x})^3-1]}[/tex]
[tex]\Leftrightarrow A= \frac{\sqrt{x}(\sqrt{x}-2)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}+\frac{\sqrt{x}+1}{\sqrt{x}(x+\sqrt{x}+1)}+\frac{1+2x-2\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}[/tex]
[tex]\Leftrightarrow A=\frac{x(\sqrt{x}-2)+(\sqrt{x}+1)(\sqrt{x}-1)+1+2x-2\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}[/tex]
[tex]\Leftrightarrow A= \frac{x\sqrt{x}-2x+x-1+1+2x-2\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}[/tex]
[tex]\Leftrightarrow A=\frac{x\sqrt{x}+x-2\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)} = \frac{\sqrt{x}(x+\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}[/tex]
Vậy A = [tex]\frac{x+\sqrt{x}-2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}[/tex]
Câu b, mk chưa nghĩ ra ,bạn thông cảm nha