Question

[Hải Dương]

Trục căn thức ở mẫu

1) \(\frac{2}{\sqrt{20}}\)

2) \(\frac{4}{\sqrt{8}}\)

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

4) \(\frac{1}{\sqrt{6}-2}\)

5) \(\frac{1}{\sqrt{2}-\sqrt{3}}\)

6) \(\frac{9a-b}{3\sqrt{a}-\sqrt{b}}\) ( a> 0, b> 0)

7) \(\frac{3\sqrt{2}}{1-\sqrt{2}-\sqrt{3}}\)

8) \(\frac{3\sqrt{3}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\)

9) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

Answer 1

[tthnew]

Em thử nhá, ko chắc đâu

1) \(\frac{2}{\sqrt{20}}=\frac{2\sqrt{20}}{20}\) 2) \(\frac{4}{\sqrt{8}}=\frac{4\sqrt{8}}{8}\)

3) \(\frac{2+\sqrt{3}}{\sqrt{2}}=\frac{2\sqrt{2}+\sqrt{6}}{2}\) 4) \(\frac{1}{\sqrt{6}-2}=\frac{\sqrt{6}+2}{6-4}=\frac{\sqrt{6}+2}{2}\)

5) \(\frac{1}{\sqrt{2}-\sqrt{3}}=\frac{\sqrt{2}+\sqrt{3}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}=-\left(\sqrt{2}+\sqrt{3}\right)\)

6) \(\frac{9a-b}{3\sqrt{a}-\sqrt{b}}=\frac{\left(9a-b\right)\left(3\sqrt{a}+b\right)}{\left(3\sqrt{a}-\sqrt{b}\right)\left(3\sqrt{a}+\sqrt{b}\right)}=\left(3\sqrt{a}+b\right)\)

7) + 8) em chưa nghĩ ra

Answer 2

[svtkvtm]

ong tth :v

\(\frac{2}{\sqrt{20}}=\frac{\sqrt{4}}{\sqrt{4}.\sqrt{5}}=\frac{1}{\sqrt{5}}\)

\(\frac{4}{\sqrt{8}}=\frac{\sqrt{16}}{\sqrt{8}}=\sqrt{2}\)

\(\frac{2+\sqrt{3}}{\sqrt{2}}=\sqrt{2}+\frac{\sqrt{3}}{\sqrt{2}}=\sqrt{2}+\sqrt{1,5}\)

\(\frac{1}{\sqrt{6}-2}=\frac{\sqrt{6}+2}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}=\frac{\sqrt{6}+2}{2}\)

\(\frac{1}{\sqrt{2}-\sqrt{3}}=\frac{\sqrt{3}+\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{3}\right)}=\frac{\sqrt{3}+\sqrt{2}}{-1}=-\sqrt{3}-\sqrt{2}\)

7: chưa

8: chưa

9:\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\left(2+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)