rút gọn:
a) 121.75.130.169
__________________
39.60.11.198
b) 2^4.5^2.11^2.7
________________
2^3.5^3.7^2.11
c) 5^11.7^12+5^11.7^11
_______________________
5^12.7^11+9.5^11.7^11
d) 5^2.6^11.16^2+6^2.12^6.15^2
_______________________________
2.6^12.10^4-81^2.960^3
e) 25^28+25^24+…+25^24+1
25^30+25^28+…+25^2+1
Đáp án:
$\begin{array}{l}
a)\dfrac{{121.75.130.169}}{{39.60.11.198}}\\
= \dfrac{{11.11.15.5.13.10.13.13}}{{3.13.4.15.11.11.18}}\\
= \dfrac{{{{11}^2}{{.13}^3}.50.15}}{{{{11}^2}.13.15.12.18}}\\
= \dfrac{{{{13}^2}.25}}{{6.18}}\\
= \dfrac{{4225}}{{108}}\\
b)\dfrac{{{2^4}{{.5}^2}{{.11}^2}.7}}{{{2^3}{{.5}^3}{{.7}^2}.11}}\\
= \dfrac{{2.11}}{{5.7}}\\
= \dfrac{{22}}{{35}}\\
c)\dfrac{{{5^{11}}{{.7}^{12}} + {5^{11}}{{.7}^{11}}}}{{{5^{12}}{{.7}^{11}} + {{9.5}^{11}}{{.7}^{11}}}}\\
= \dfrac{{{5^{11}}{{.7}^{11}}\left( {7 + 1} \right)}}{{{5^{11}}{{.7}^{11}}\left( {5 + 9} \right)}}\\
= \dfrac{8}{{14}}\\
= \dfrac{4}{7}\\
d)\dfrac{{{5^2}{{.6}^{11}}{{.16}^2} + {6^2}{{.12}^6}{{.15}^2}}}{{{{2.6}^{12}}{{.10}^4} – {{81}^2}{{.960}^3}}}\\
= \dfrac{{{5^2}{{.2}^{11}}{{.3}^{11}}{{.2}^8} + {3^2}{{.2}^2}{{.2}^{12}}{{.3}^6}{{.3}^2}{{.5}^2}}}{{{{2.2}^{12}}{{.3}^{12}}{{.2}^4}{{.5}^4} – {3^8}.{{\left( {{2^5}.3.5} \right)}^3}}}\\
= \dfrac{{{5^2}{{.2}^{19}}{{.3}^{11}} + {5^2}{{.2}^{14}}{{.3}^{10}}}}{{{5^2}{{.2}^{18}}{{.3}^{12}} – {5^3}{{.2}^{15}}{{.3}^{11}}}}\\
= \dfrac{{{5^2}{{.2}^{14}}{{.3}^{10}}\left( {{2^5} + 1} \right)}}{{{5^2}{{.2}^{15}}{{.3}^{11}}\left( {{2^3}.3 – 5} \right)}}\\
= \dfrac{{33}}{{2.3.19}}\\
= \dfrac{{11}}{{38}}\\
e)A = {25^{28}} + {25^{24}} + … + {25^4} + 1\\
\Rightarrow {25^4}.A = {25^{32}} + {25^{28}} + … + {25^8} + {25^4}\\
\Rightarrow {25^4}A – A = {25^{32}} – 1\\
\Rightarrow A = \dfrac{{{{25}^{32}} – 1}}{{{{25}^4} – 1}}\\
B = {25^{30}} + {25^{28}} + … + {25^2} + 1\\
\Rightarrow {25^2}.B = {25^{32}} + {25^{30}} + … + {25^4} + {25^2}\\
\Rightarrow {25^2}B – B = {25^{32}} – 1\\
\Rightarrow B = \dfrac{{{{25}^{32}} – 1}}{{{{25}^2} – 1}}
\end{array}$