Câu 11:

Tính:

a) \(\sqrt{16} \cdot \sqrt{25}+\sqrt{196}: \sqrt{49}\)

b) \(36: \sqrt{2.3^{2} .18}-\sqrt{169}\)

c) \(\sqrt{\sqrt{81}}\)

d) \(\sqrt{3^{2}+4^{2}}\)

Lời giải:

a) \(\sqrt{16} \cdot \sqrt{25}+\sqrt{196}: \sqrt{49}\) = \(\sqrt{4^{2}} \cdot \sqrt{5^{2}}+\sqrt{14^{2}}: \sqrt{7^{2}}\)

= 4.5 + 14:7 = 20 + 2 = 22

b) \(36: \sqrt{2.3^{2} .18}-\sqrt{169}\) = \(36: \sqrt{3^{2} \cdot 36}-\sqrt{13^{2}}\)

= \(36: \sqrt{3^{2} \cdot 6^{2}}-13\) = \(36: \sqrt{18^{2}}-13\) = 36 : 18 – 13 = - 11

c) \(\sqrt{\sqrt{81}}\) = \(\sqrt{\sqrt{9^{2}}}=\sqrt{9}=\sqrt{3^{2}}=3\)

d) \(\sqrt{3^{2}+4^{2}}\) = \(\sqrt{9+16}=\sqrt{25}=\sqrt{5^{2}}=5\)