Bài 29 (trang 19 SGK Toán 9 Tập 1) Tính a) $\dfrac{\sqrt{2}}{\sqrt{18}}$ ;                             b) $\dfrac{\sqrt{15}}{\sqrt{735}}$ ; c) $\dfrac{\sqrt{12500}}{\sqrt{500}}$ ;                       d) $\dfrac{\sqrt{6^5}}{\sqrt{2^3.3^5}}$.
Help me, please! Mình đang tìm kiếm câu trả lời cho một câu hỏi cực kỳ khó khăn và mình cần sự trợ giúp từ cộng đồng. Ai có thể giải đáp giúp mình?

Để giải bài toán trên, ta sử dụng tính chất của căn bậc 2:

a) $\dfrac{\sqrt{2}}{\sqrt{18}} = \dfrac{\sqrt{2}}{\sqrt{2^2\cdot 3}} = \sqrt{\dfrac{2}{2^2\cdot 3}} = \sqrt{\dfrac{1}{2\cdot 3}} = \sqrt{\dfrac{1}{6}} = \dfrac{1}{\sqrt{6}} = \dfrac{\sqrt{6}}{6}$

b) $\dfrac{\sqrt{15}}{\sqrt{735}} = \dfrac{\sqrt{3\cdot 5}}{\sqrt{3\cdot 5\cdot 7}} = \dfrac{\sqrt{3\cdot 5}}{\sqrt{3\cdot 5\cdot 7}} = \sqrt{\dfrac{3\cdot 5}{3\cdot 5\cdot 7}} = \sqrt{\dfrac{1}{7}} = \dfrac{1}{\sqrt{7}} = \dfrac{\sqrt{7}}{7}$

c) $\dfrac{\sqrt{12500}}{\sqrt{500}} = \dfrac{\sqrt{5^4\cdot 2^2}}{\sqrt{5^3\cdot 2^2}} = \sqrt{\dfrac{5^4\cdot 2^2}{5^3\cdot 2^2}} = \sqrt{\dfrac{5}{5}} = 1$

d) $\dfrac{\sqrt{6^5}}{\sqrt{2^3\cdot 3^5}} = \dfrac{6^{\frac{5}{2}}}{2^{\frac{3}{2}}\cdot 3^{\frac{5}{2}}} = \dfrac{6^{\frac{5}{2}}}{6^{\frac{3}{2}}\cdot 3^{\frac{5}{2}}} = \dfrac{1}{6^{\frac{1}{2}}\cdot 3^{\frac{1}{2}}} = \sqrt{\dfrac{1}{6\cdot 3}} = \sqrt{\dfrac{1}{18}} = \dfrac{1}{\sqrt{18}} = \dfrac{\sqrt{18}}{18}$

Vậy kết quả của các phép tính là:

a) $\dfrac{\sqrt{2}}{\sqrt{18}} = \dfrac{\sqrt{6}}{6}$

b) $\dfrac{\sqrt{15}}{\sqrt{735}} = \dfrac{\sqrt{7}}{7}$

c) $\dfrac{\sqrt{12500}}{\sqrt{500}} = 1$

d) $\dfrac{\sqrt{6^5}}{\sqrt{2^3\cdot 3^5}} = \dfrac{\sqrt{18}}{18}$

{
"content1": {
"a": "a) $\dfrac{\sqrt{2}}{\sqrt{18}} = \dfrac{\sqrt{2}}{\sqrt{2^2 \times 3}} = \dfrac{\sqrt{2}}{2\sqrt{3}} = \dfrac{1}{2\sqrt{3}} = \dfrac{\sqrt{3}}{2 \times 3} = \dfrac{\sqrt{3}}{6}$",
"b": "$\dfrac{\sqrt{15}}{\sqrt{735}} = \dfrac{\sqrt{3 \times 5}}{\sqrt{3^2 \times 5 \times 7}} = \dfrac{\sqrt{3} \times \sqrt{5}}{3\sqrt{5 \times 7}} = \dfrac{\sqrt{3}}{3\sqrt{7}} = \dfrac{\sqrt{3}}{3\sqrt{7}}$",
"c": "$\dfrac{\sqrt{12500}}{\sqrt{500}} = \dfrac{\sqrt{5^4 \times 5^2}}{\sqrt{5^2 \times 2^2 \times 5^2}} = \dfrac{5^2 \times \sqrt{5^2}}{5 \times 2 \times 5} = \dfrac{5 \times 5}{5 \times 2 \times 5} = \dfrac{1}{2}$",
"d": "$\dfrac{\sqrt{6^5}}{\sqrt{2^3 \times 3^5}} = \dfrac{\sqrt{6^4 \times 6}}{\sqrt{2^3 \times 3^4 \times 3}} = \dfrac{6^2 \times \sqrt{6}}{2 \times 3^2 \times \sqrt{3}} = \dfrac{36 \times \sqrt{6}}{2 \times 9 \times \sqrt{3}} = \dfrac{18 \times \sqrt{6}}{18 \times \sqrt{3}} = \dfrac{\sqrt{6}}{\sqrt{3}} = \sqrt{2}$"
},
"content2": {
"a": "a) $\dfrac{\sqrt{2}}{\sqrt{18}} = \sqrt{\dfrac{2}{18}} = \sqrt{\dfrac{1}{9}} = \dfrac{1}{3}$",
"b": "$\dfrac{\sqrt{15}}{\sqrt{735}} = \sqrt{\dfrac{15}{735}} = \sqrt{\dfrac{1}{49}} = \dfrac{1}{7}$",
"c": "$\dfrac{\sqrt{12500}}{\sqrt{500}} = \sqrt{\dfrac{12500}{500}} = \sqrt{25} = 5$",
"d": "$\dfrac{\sqrt{6^5}}{\sqrt{2^3 \times 3^5}} = \sqrt{\dfrac{6^5}{2^3 \times 3^5}} = \sqrt{\dfrac{2^5 \times 3^5}{2^3 \times 3^5}} = \sqrt{\dfrac{2^2}{2^3}} = \dfrac{1}{2}$"
},
"content3": {
"a": "a) $\dfrac{\sqrt{2}}{\sqrt{18}} = \sqrt{\dfrac{2}{18}} = \sqrt{\dfrac{2}{2^2 \times 3}} = \sqrt{\dfrac{1}{2 \times 3}} = \dfrac{1}{\sqrt{6}} = \dfrac{1}{2\sqrt{6}} = \dfrac{\sqrt{6}}{12}$",
"b": "$\dfrac{\sqrt{15}}{\sqrt{735}} = \sqrt{\dfrac{15}{735}} = \sqrt{\dfrac{15}{3 \times 5 \times 7 \times 7}} = \sqrt{\dfrac{1}{3 \times 7^2}} = \dfrac{1}{7\sqrt{3}}$",
"c": "$\dfrac{\sqrt{12500}}{\sqrt{500}} = \sqrt{\dfrac{12500}{500}} = \sqrt{25} = 5$",
"d": "$\dfrac{\sqrt{6^5}}{\sqrt{2^3 \times 3^5}} = \sqrt{\dfrac{6^5}{2^3 \times 3^5}} = \sqrt{\dfrac{2^5 \times 3^5}{2^3 \times 3^5}} = \sqrt{\dfrac{2^2}{2^3}} = \dfrac{\sqrt{2}}{\sqrt{8}} = \dfrac{\sqrt{2}}{2\sqrt{2}} = \dfrac{1}{2}$"
}
}