Question

1) \(\sqrt{9a^2.b^2}\) với a<0, b<0

2) \(\sqrt{3a}.\sqrt{27a}\) với a \(\ge\)0

3) \(\sqrt{3a^5}.12a\) với a>0

4) \(\sqrt{5a}.\sqrt{45a}-3a\) ( với a ≥  0)

5) \(\sqrt{3+\sqrt{a}}\).\(\sqrt{3-\sqrt{a}}\)

6) \(\sqrt{3+\sqrt{5}}\). \(\sqrt{3\sqrt{5}}\)

 

Answer 1

\(1) \sqrt{9a^2.b^2}\)=3ab

\(2) \sqrt{3a}.\sqrt{27a}=\sqrt{3a}.3\sqrt{3a}=9a\)

\(3) \sqrt{3a^5}.12a=12\sqrt{3a^7}\)

\(4) \sqrt{5a}.\sqrt{45a}-3a=15a-3a=12a\)

\(5) \sqrt{3+\sqrt{a}}.\sqrt{3-\sqrt{a}}=\sqrt{(3+\sqrt{a}).(3-\sqrt{a})} =\sqrt{9-a} \)

\(6) \sqrt{3+\sqrt{5}}.\sqrt{3\sqrt{5}} =\sqrt{\sqrt{3\sqrt{5}}.(3+\sqrt{5})} =\sqrt{9+\sqrt{15}}\)

 

Answer 2

1) \(\sqrt{9a^2b^2}=3ab\)

2) \(\sqrt{3a}\cdot\sqrt{27a}=9a\)

4) \(\sqrt{5a}\cdot\sqrt{45a}-3a=15a-3a=12a\)