Đáp án:
$\dfrac{3^6.21^{12}}{175^9.7^3}<\dfrac{3^{10}.6^7.4}{10^9.5^8}.$
Giải thích các bước giải:
$\dfrac{3^6.21^{12}}{175^9.7^3}\\ =\dfrac{3^6.(3.7)^{12}}{(5^2.7)^9.7^3}\\ =\dfrac{3^6.3^{12}.7^{12}}{5^{18}.7^9.7^3}\\ =\dfrac{3^{18}.7^{12}}{5^{18}.7^{12}}\\ =\dfrac{3^{18}}{5^{18}}\\ =\left(\dfrac{3}{5}\right)^{18}\\ \dfrac{3^{10}.6^7.4}{10^9.5^8}\\ =\dfrac{3^{10}.(2.3)^7.2^2}{(2.5)^9.5^8}\\ =\dfrac{3^{10}.2^7.3^7.2^2}{2^9.5^9.5^8}\\ =\dfrac{3^{17}.2^9}{2^9.5^{17}}\\ =\dfrac{3^{17}}{5^{17}}\\ =\left(\dfrac{3}{5}\right)^{17}\\ 0<\dfrac{3}{5}<1 \Rightarrow \left(\dfrac{3}{5}\right)^{18}<\left(\dfrac{3}{5}\right)^{17} \Leftrightarrow \dfrac{3^6.21^{12}}{175^9.7^3}<\dfrac{3^{10}.6^7.4}{10^9.5^8}.%$
