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Đáp án:
c,
$\frac{8018}{2004.2006-2003.2005}$
=$\frac{8018}{(2005-1)(2005+1)-(2004-1).(2004+1)}$
=$\frac{8018}{2005^{2}-1-2004^{2}+1}$
=$\frac{8018}{(2005-2004)(2005+2004)}$
=$\frac{8018}{4009}$ = 2
a,
$\frac{5.4^{15}.9^{9}-4.3^{20}.8^{9}}{5.2^{9}.6^{19}-7.2^{29}.27^{6}}$
Xét tử số:
$5.4^{15}.9^{9}-4.3^{20}.8^{9}$
= $5.2^{30}.3^{18}-3^{20}.2^{29}$
= $3^{18}.2^{29}.(5.2-3^{2})$
= $3^{18}.2^{29}$
Xét mẫu số:
$5.2^{9}.6^{19}-7.2^{29}.27^{6}$
= $5.2^{9}.(2.3)^{19}-7.2^{29}.(3^{3})^{6}$
= $5.2^{28}.3^{19}-7.2^{29}.3^{18}$
= $2^{28}.3^{18}.(5.3-7.2)$
= $2^{28}.3^{18}$
=> $\frac{5.4^{15}.9^{9}-4.3^{20}.8^{9}}{5.2^{9}.6^{19}-7.2^{29}.27^{6}}$ $(2^{3})^{6}$
= $\frac{3^{18}.2^{29}}{2^{28}.3^{18}}$
= 2
b,
Xét tử số:
$35(27^{8}+2.9^{11})$
= $35((3^{3})^{8}+2.(3^{2})^{11})$
= $35(3^{24}+2.3^{22})$
= $35.3^{22}.(3^{2}+2)$
= $35.11.3^{22}$
Xét mẫu số:
$15(81^{6}-12.3^{19})$
= $15((3^{4})^{6}-12.3^{19})$
= $15(3^{24}-12.3^{19})$
= $15.3^{19}.(3^{5}-12)$
= $15.231.3^{19}$
=>$\frac{35(27^{8}+2.9^{11})}{15(81^{6}-12.3^{19})}$
= $\frac{35.11.3^{22}}{15.231.3^{19}}$
= $\frac{3^{3}}{9}$
= 3
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