Question

[Nguyễn Minh Hùng]

Chỉ giúp em với ạ bài 31

Answer 1

[An Thy]

1. \(\dfrac{2}{2-\sqrt{3}}=\dfrac{2\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\dfrac{4+2\sqrt{3}}{2^2-\left(\sqrt{3}\right)^2}=\dfrac{4+2\sqrt{3}}{4-3}=4+2\sqrt{3}\)

2. \(\dfrac{1}{\sqrt{3}+\sqrt{2}}=\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}=\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2}=\dfrac{\sqrt{3}-\sqrt{2}}{3-2}\)

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

3. \(\dfrac{1}{\sqrt{5}+\sqrt{7}}=\dfrac{\sqrt{7}-\sqrt{5}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)}=\dfrac{\sqrt{7}-\sqrt{5}}{\left(\sqrt{7}\right)^2-\left(\sqrt{5}\right)^2}=\dfrac{\sqrt{7}-\sqrt{5}}{7-5}\)

\(=\dfrac{\sqrt{7}-\sqrt{5}}{2}\)

4. \(\dfrac{1}{5-2\sqrt{6}}=\dfrac{5+2\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}=\dfrac{5+2\sqrt{6}}{5^2-\left(2\sqrt{6}\right)^2}=\dfrac{5+2\sqrt{6}}{25-24}\)

\(=5+2\sqrt{6}\)

5. \(\dfrac{3\sqrt{5}}{2\sqrt{5}-1}=\dfrac{3\sqrt{5}\left(2\sqrt{5}+1\right)}{\left(2\sqrt{5}-1\right)\left(2\sqrt{5}\right)+1}=\dfrac{30+3\sqrt{5}}{\left(2\sqrt{5}\right)^2-1^2}=\dfrac{30+3\sqrt{5}}{20-1}\)

\(=\dfrac{30+3\sqrt{5}}{19}\)

6. \(\dfrac{12}{3-\sqrt{3}}=\dfrac{12}{\sqrt{3}\left(\sqrt{3}-1\right)}=\dfrac{4\sqrt{3}}{\sqrt{3}-1}=\dfrac{4\sqrt{3}\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(\dfrac{12+4\sqrt{3}}{\left(\sqrt{3}\right)^2-1^2}=\dfrac{2\left(6+2\sqrt{3}\right)}{3-1}=6+2\sqrt{3}\)

7. \(\dfrac{5\sqrt{2}}{\sqrt{5}+\sqrt{3}}=\dfrac{5\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\dfrac{5\sqrt{10}-5\sqrt{6}}{\left(\sqrt{5}\right)^2-\left(\sqrt{3}\right)^2}\)

\(=\dfrac{5\sqrt{10}-5\sqrt{6}}{5-3}=\dfrac{5\sqrt{10}-5\sqrt{6}}{2}\)

8. \(\dfrac{18}{\sqrt{7}-1}=\dfrac{18\left(\sqrt{7}+1\right)}{\left(\sqrt{7}-1\right)\left(\sqrt{7}+1\right)}=\dfrac{18\left(\sqrt{7}+1\right)}{\left(\sqrt{7}\right)^2-1^2}=\dfrac{18\left(\sqrt{7}+1\right)}{7-1}\)

\(=3\left(\sqrt{7}+1\right)=3\sqrt{7}+3\)

9. \(\dfrac{9}{2\sqrt{3}-3}=\dfrac{9\left(2\sqrt{3}+3\right)}{\left(2\sqrt{3}-3\right)\left(2\sqrt{3}+3\right)}=\dfrac{9\left(2\sqrt{3}+3\right)}{\left(2\sqrt{3}\right)^2-3^2}=\dfrac{9\left(2\sqrt{3}+3\right)}{12-9}\)

\(3\left(2\sqrt{3}+3\right)=6\sqrt{3}+9\)

10. \(\dfrac{1}{2\sqrt{3}-3}=\dfrac{2\sqrt{3}+3}{\left(2\sqrt{3}-3\right)\left(2\sqrt{3}+3\right)}=\dfrac{2\sqrt{3}+3}{\left(2\sqrt{3}\right)^2-3^2}=\dfrac{2\sqrt{3}+3}{12-9}\)

\(=\dfrac{2\sqrt{3}+3}{3}\)

11. \(\dfrac{3}{2\sqrt{2}-\sqrt{5}}=\dfrac{3\left(2\sqrt{2}+\sqrt{5}\right)}{\left(2\sqrt{2}-\sqrt{5}\right)\left(2\sqrt{2}+\sqrt{5}\right)}=\dfrac{3\left(2\sqrt{2}+\sqrt{5}\right)}{\left(2\sqrt{2}\right)^2-\left(\sqrt{5}\right)^2}\)

\(=\dfrac{3\left(2\sqrt{2}+\sqrt{5}\right)}{8-5}=2\sqrt{2}+5\)

12. \(\dfrac{1+\sqrt{2}}{1-\sqrt{2}}=\dfrac{\left(1+\sqrt{2}\right)\left(1+\sqrt{2}\right)}{\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}=\dfrac{\left(1+\sqrt{2}\right)^2}{1^2-\left(\sqrt{2}\right)^2}=\dfrac{3+2\sqrt{2}}{-1}\)

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

13. \(\dfrac{\sqrt{3}+2}{2-\sqrt{3}}=\dfrac{\left(\sqrt{3}+2\right)\left(\sqrt{3}+2\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\dfrac{\left(\sqrt{3}+2\right)^2}{2^2-\left(\sqrt{3}\right)^2}=\dfrac{7+4\sqrt{3}}{4-3}=7+4\sqrt{3}\)

14. \(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}=\dfrac{\left(3+\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}=\dfrac{\left(3+\sqrt{5}\right)^2}{3^2-\left(\sqrt{5}\right)^2}=\dfrac{14+6\sqrt{5}}{9-5}\)

\(=\dfrac{7+3\sqrt{5}}{2}\)

15. giống câu 5

16. \(\dfrac{\sqrt{5}+1}{2\sqrt{5}-4}=\dfrac{\left(\sqrt{5}+1\right)\left(2\sqrt{5}+4\right)}{\left(2\sqrt{5}-4\right)\left(2\sqrt{5}+4\right)}=\dfrac{14+6\sqrt{5}}{\left(2\sqrt{5}\right)^2-4^2}=\dfrac{14+6\sqrt{5}}{4}\)

\(=\dfrac{7+3\sqrt{5}}{2}\)

Answer 2

[Nguyễn Ngọc Lộc]

- Sử dụng liên hợp thôi nha mình làm tham khảo câu 1, 4 các câu khác tương tự .

\(1,\dfrac{2}{2-\sqrt{3}}=\dfrac{2\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\dfrac{4+2\sqrt{3}}{4-3}=3+2\sqrt{3}+1=\left(\sqrt{3}+1\right)^2\)

\(4,\dfrac{1}{5-2\sqrt{6}}=\dfrac{5+2\sqrt{6}}{5^2-\left(2\sqrt{6}\right)^2}=5+2\sqrt{6}\)