Question

[Loveduda]

R/gọn và tìm điều kiện xác định:
\(M=\left(\dfrac{2}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x+1}}\right).\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)

Answer 1

[Đặng Quý]

sửa đề:

\(M=\left(\dfrac{2}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)

ĐKXĐ: \(x\ge0\);\(x\ne\pm1\)

\(M=\left(\dfrac{2}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\\M=\left[\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right].\dfrac{\sqrt{x}}{x+\sqrt{x}+2} \)

\(M=\dfrac{x+\sqrt{x}+2}{x-1}.\dfrac{\sqrt{x}}{x+\sqrt{x}+2}=\dfrac{\sqrt{x}}{x-1}\)