Question

[Việt Lê]

1) Rút gọn : A=\(\frac{\sqrt{8-2\sqrt{15}}}{\sqrt{10}-\sqrt{6}}\)

2) Rút gọn : B= \(\left(\frac{\sqrt{a}}{\sqrt{a-2}}+\frac{\sqrt{a}}{\sqrt{a+2}}\right)\): \(\frac{\sqrt{4a}}{\sqrt{a-4}}\)

(a>0 ; a ≠ 4)

3) Chứng minh rằng

\(\left(\frac{1}{\sqrt{1+a}}\sqrt{1-a}\right):\left(\frac{1}{\sqrt{1-a^2}}\right)=\sqrt{1-a}\)

Điều kiện (-1<a<1)

Hóng cao nhân giải bài này ???

Answer 1

[Hoàng Anh Thư]

câu 3???? sai???

Answer 2

[Y]

1. \(A=\frac{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)}=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)}=\frac{1}{\sqrt{2}}\)

3. \(\frac{\sqrt{1-a}}{\sqrt{1+a}}:\frac{1}{\sqrt{1-a^2}}\) \(=\frac{\sqrt{\left(1-a\right)}\cdot\sqrt{1-a}}{\sqrt{1+a}\cdot\sqrt{1-a}}\cdot\sqrt{1-a^2}\)

\(=\frac{1-a}{\sqrt{1-a^2}}\cdot\sqrt{1-a^2}=1-a\)