Question

a, \(\sqrt{9\cdot\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}\)

b,\(\sqrt{9\left(3-a\right)^2}vớia>3\)

 

Answer 1

a) \(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{81-17}=\sqrt{64}=8\)

b)\(\sqrt{9\left(3-a\right)^2}=3\left|3-a\right|=3\left(a-3\right)\)(vì a > 3)

Answer 2

\(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}\)

\(=\sqrt{\left(\sqrt{9}\right)^2}-\sqrt{\left(\sqrt{17}\right)^2}\)

\(\sqrt{9\left(3-a\right)^2}\)

\(=\sqrt{3^2\left(3-a\right)^2}\)

\(=3\left(3-a\right)\)

\(=3-3a\)