Question

\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}2-\dfrac{2+5\sqrt{x}}{x-4},x\ge0,x\ne4\)

Answer 1

Lời giải:

Nếu yêu cầu đề là rút gọn thì giải như sau:

\(BT = \frac{(\sqrt{x}+1)(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{2\sqrt{x}(\sqrt{x}-2)}{(\sqrt{x}+2)(\sqrt{x}-2)}-\frac{2+5\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\)

\(=\frac{(x+3\sqrt{x}+2)+(2x-4\sqrt{x})-(2+5\sqrt{x})}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{3x-6\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{3\sqrt{x}(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{3\sqrt{x}}{\sqrt{x}+2}\)

Answer 2

Số `2` đó là `*2` hay sao bạn?