Câu 29:

Tính:

a) \(\frac{\sqrt{2}}{\sqrt{18}}\)

b) \(\frac{\sqrt{15}}{\sqrt{735}}\)

c) \(\frac{\sqrt{12500}}{\sqrt{500}}\)

d) \(\frac{\sqrt{6^{5}}}{\sqrt{2^{3} \cdot 3^{5}}}\)

Lời giải:

a) \(\frac{\sqrt{2}}{\sqrt{18}}=\sqrt{\frac{2}{18}}=\sqrt{\frac{1}{9}}=\frac{\sqrt{1}}{\sqrt{9}}=\frac{1}{3}\)

b) \(\frac{\sqrt{15}}{\sqrt{735}}=\sqrt{\frac{15}{735}}=\sqrt{\frac{1}{49}}=\frac{\sqrt{1}}{\sqrt{49}}=\frac{1}{7}\)

c) \(\frac{\sqrt{12500}}{\sqrt{500}}=\sqrt{\frac{12500}{500}}=\sqrt{25}=5\)

d) \(\frac{\sqrt{6^{5}}}{\sqrt{2^{3} \cdot 3^{5}}}=\sqrt{\frac{6^{5}}{2^{3} \cdot 3^{5}}}=\sqrt{\frac{(2 \cdot 3)^{5}}{2^{3} \cdot 3^{5}}}=\sqrt{\frac{2^{5} \cdot 3^{5}}{2^{3} \cdot 3^{5}}}=\sqrt{2^{2}}=2\)