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

Question

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= $\frac{2}{x+\sqrt[]{x}+1}$

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1 giờ 2021-09-07T10:40:26+00:00 2 Answers 0 views 0

1. Tham khảo!
Ảnh 1: Tìm ĐKXĐ

Ảnh 2: Rút gọn.

2. 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 = $\frac{2}{x+√x+1}$