Question

Answer 1

Bạn cần bài nào thì chụp/ gõ nguyên bài đó ra.

Answer 2

Bạn cần bài nào vậy?

Answer 3

3.

\(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}=\frac{\sqrt{10}(\sqrt{10}+2)}{\sqrt{5}+\sqrt{2}}+\frac{8(1+\sqrt{5})}{(1+\sqrt{5})(1-\sqrt{5})}\)

\(=\frac{\sqrt{10}.\sqrt{2}(\sqrt{5}+\sqrt{2})}{\sqrt{5}+\sqrt{2}}+\frac{8(\sqrt{5}+1)}{-4}\)

\(=\sqrt{20}+(-2)(\sqrt{5}+1)=2\sqrt{5}-2\sqrt{5}-2=-2\)

 

Answer 4

4.

\(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=\frac{2(\sqrt{8}-\sqrt{3})}{\sqrt{6}(\sqrt{3}-\sqrt{8})}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{6}(\sqrt{5}+\sqrt{27})}\)

\(=\frac{-2}{\sqrt{6}}-\frac{1}{\sqrt{6}}=\frac{-3}{\sqrt{6}}=\frac{-\sqrt{6}}{2}\)

Answer 5

5.

\(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}=8\sqrt{\frac{1}{3}}-\sqrt{\frac{1}{3}}-\frac{12}{5}\sqrt{\frac{1}{3}}\)

\(=(8-1-\frac{12}{5}).\sqrt{\frac{1}{3}}=\frac{23}{5}\sqrt{\frac{1}{3}}=\frac{23\sqrt{3}}{15}\)

 

Answer 6

6.

\(\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}=\frac{(2-\sqrt{3})+(2+\sqrt{3})}{\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}\)

\(=\frac{4}{\sqrt{2^2-3}}=\frac{4}{1}=4\)

Answer 7

7.

\(2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}\)

\(=2.3\sqrt{3}-\frac{12}{\sqrt{3}}+3\sqrt{3}=6\sqrt{3}-4\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)