2. Chứng minh $B=\frac{\sqrt{x}+1}{\sqrt{x}-1}$;

Q: 2. Chứng minh $B=\frac{\sqrt{x}+1}{\sqrt{x}-1}$;

<p><span class="math-tex">$\begin{array}{l}\text { Xét } B=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-1}=\frac{\sqrt{x}-1+\sqrt{x}(\sqrt{x}+1)+2}{x-1}=\frac{\sqrt{x}-1+x+\sqrt{x}+2}{x-1} \\ \Rightarrow B=\frac{\sqrt{x}-1+x+\sqrt{x}+2}{x-1}=\frac{(\sqrt{x}+1)^{2}}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{\sqrt{x}+1}{\sqrt{x}-1} .\end{array}$</span></p>

Question

Accepted Answer

$\begin{array}{l}\text { Xét } B=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-1}=\frac{\sqrt{x}-1+\sqrt{x}(\sqrt{x}+1)+2}{x-1}=\frac{\sqrt{x}-1+x+\sqrt{x}+2}{x-1} \ \Rightarrow B=\frac{\sqrt{x}-1+x+\sqrt{x}+2}{x-1}=\frac{(\sqrt{x}+1)^{2}}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{\sqrt{x}+1}{\sqrt{x}-1} .\end{array}$

Đề thi cuối kì 2 (CT) 20-21 - Q. Bắc Từ Liêm - Tp. Hà Nội - MĐ 5539