Giải hộ em với ạ, em cảm ơn. Question Giải hộ em với ạ, em cảm ơn. in progress 0 Môn Toán Gerda 5 years 2021-05-23T17:13:34+00:00 2021-05-23T17:13:34+00:00 2 Answers 27 views 0
Answers ( )
Đáp án:
Giải thích các bước giải:
a)
`19/(2^2 . 3^2 . 5^4)=(19 . 2^2)/(2^4 . 3^2 . 5^4)=76/(2^4 . 3^2 . 5^4)`
`11/(2^4 . 3 . 5^3)=(11 . 3 . 5)/(2^4 . 3^2 . 5^4)=165/(2^4 . 3^2 . 5^4)`
Vì `76<165 to 76/(2^4 . 3^2 . 5^4)<165/(2^4 . 3^2 . 5^4)`
hay `19/(2^2 . 3^2 . 5^4)<11/(2^4 . 3 . 5^3)`
b)
`13/(3 . 7^2 . 11)=(13 . 3 . 13)/(3^2 . 7^2 . 11 . 13)=507/(3^2 . 7^2 . 11 . 13)`
`17/(3^2 . 7 . 13)=(17 . 7 . 11)/(3^2 . 7^2 . 11 . 13)=1309/(3^2 . 7^2 . 11 . 13)`
Vì `507<1309 to 507/(3^2 . 7^2 . 11 . 13)<1309/(3^2 . 7^2 . 11 . 13)`
hay `13/(3 . 7^2 . 11)<17/(3^2 . 7 . 13)`
Giải thích các bước giải:
a) Ta có:
$\begin{array}{l}
\dfrac{{19}}{{{2^2}{{.3}^2}{{.5}^4}}} = \dfrac{{{{19.2}^2}}}{{{2^4}{{.3}^2}{{.5}^4}}} = \dfrac{{76}}{{{2^4}{{.3}^2}{{.5}^4}}}\\
\dfrac{{11}}{{{2^4}{{.3.5}^3}}} = \dfrac{{11.3.5}}{{{2^4}{{.3}^2}{{.5}^4}}} = \dfrac{{165}}{{{2^4}{{.3}^2}{{.5}^4}}}
\end{array}$
b) Ta có:
$\begin{array}{l}
\dfrac{{13}}{{{{3.7}^2}.11}} = \dfrac{{13.3.13}}{{{3^2}{{.7}^2}.11.13}} = \dfrac{{507}}{{{3^2}{{.7}^2}.11.13}}\\
\dfrac{{17}}{{{3^2}.7.13}} = \dfrac{{17.7.11}}{{{3^2}{{.7}^2}.11.13}} = \dfrac{{1309}}{{{3^2}{{.7}^2}.11.13}}
\end{array}$