Cho
A= $\frac{x+2}{x\sqrt[]{x}-1}$ + $\frac{\sqrt[]{x}}{x+\sqrt[]{x}+1}$ + $\frac{1}{1-\sqrt[]{x}}$ : $\frac{\sqrt[]{x}-1}{2 }$
a, Tìm ĐKXĐ
b, Chứng minh A= $\frac{2}{x+\sqrt[]{x}+1}$
Cho A= $\frac{x+2}{x\sqrt[]{x}-1}$ + $\frac{\sqrt[]{x}}{x+\sqrt[]{x}+1}$ + $\frac{1}{1-\sqrt[]{x}}$ : $\frac{\sqrt[]{x}-1}{2 }$ a, Tìm ĐKXĐ b, Chứn
By Liliana
Tham khảo!
Ảnh 1: Tìm ĐKXĐ
Ảnh 2: Rút gọn.
A = ($\frac{x+2}{x√x-1}$ + $\frac{√x}{x+√x+1}$ + $\frac{1}{1-√x}$) : $\frac{√x-1}{2}$
(ĐKXĐ: x $\neq$ 1; x ≥ 0)
A = [$\frac{x+2}{(√x-1)(x+√x+1)}$ + $\frac{√x}{x+√x+1}$ – $\frac{1}{√x-1}$] . $\frac{2}{√x-1}$
A = $\frac{x+2+x-√x-x-√x-1}{(√x-1)(x+√x+1)}$ . $\frac{2}{√x-1}$
A = $\frac{(√x-1)²}{(√x-1)(x+√x+1)}$ . $\frac{2}{√x-1}$
A = $\frac{2}{x+√x+1}$