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Ta có: 76 + 75 - 74
= 74 . (49+7-1)
= 74 . 55 chia hết cho 11 => ĐPCM
Ta có: 2454⋅5424⋅210
= (23 . 3)54 . (33 . 2) . 210
= 2162 . 354 . 372. 224 . 210
= 2196 . 3126
= (2189 . 3126). 27
=7263 . 27 chia hết cho 63 => ĐPCM
77^6+7^5-7^4
=7^6.11^6+7^5-7^4
=7^4.7^2+7^4.7-7^4.1.11^6
=7^4.(7^2+7-1).11^6 chia hết cho 7
77^6+7^5-7^4 chia hết vì có số 7^4=7.7^3
Ta có 2454.5424.210=(23.3)54.(33.2)24.210=2162.354.372.224.210=2196.3126=(2189.3126).27=7263.27chia hết cho 7263(vì 7263chia hết cho 7263) => đpcm
\(7^6+7^5-7^4=7^4\left(49+7-1\right)=7^4.55\)chia hết cho 11
=>ĐPCM
\(24^{54}.54^{24}.2^{10}=8^{54}.3^{54}.27^{24}.2^{24}.2^{10}=2^{162}.3^{54}.3^{72}.2^{34}\)
\(=2^{196}.3^{126}=2^{189}.2^7.9^{63}=8^{63}.9^{63}.2^7=72^{63}.2^7\)chia hết cho 7263
=>đpcm
\(7^6+7^5-7^4\)
\(=7^4\cdot7^2+7^5\cdot7-7^4\)
\(=7^4\cdot\left(7^2+7-1\right)\)
\(=7^4\cdot55\)
\(=7^4\cdot5\cdot11⋮11\left(đpcm\right)\)
\(7^6+7^5-7^4=7^4.\left(7^2+7-1\right)\)
\(=7^4.55⋮11\)
\(=>7^6+7^5-7^4⋮11\)
a, Ta có:
\(81^7-27^9-9^{13}=3^{28}-3^{27}-3^{26}\)
\(=3^{26}\left(3^2-3-1\right)=3^{25}.3.5=3^{25}.15\)
Vì 15 chia hết cho 15 nên \(3^{25}.15\) chia hết cho 15.
Vậy................(đpcm)
b,Ta có:
\(24^{54}.54^{24}.2^{10}=\left(2^3.3\right)^{54}.\left(2.3^3\right)^{24}.2^{10}\)
\(=2^{162}.3^{54}.2^{24}.3^{72}.2^{10}=2^{196}.3^{126}\)
\(=2^{108}.3^{72}.2^{88}.3^{54}\)
\(72^{36}=\left(2^3.3^2\right)^{36}=2^{108}.3^{72}\)
Vì \(2^{108}.3^{72}\) chia hết cho \(2^{108}.3^{72}\) nên \(2^{108}.3^{72}.2^{88}.3^{54}\) chia hết cho \(2^{108}.3^{72}\)
Vậy............(đpcm)
Chúc bạn học tốt!!!
a, \(81^7-27^9-9^{13}\)
\(=3^{28}-3^{27}-3^{26}\)
\(=3^{22}\left(3^6-3^5-3^4\right)\)
\(=3^{22}\times405⋮405\)
a)
\(A=\left(\frac{19}{24}-\frac{7}{24}\right)-\left(\frac{1}{2}+\frac{1}{3}\right)\)
\(A=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}\)
\(A=\frac{1}{3}\)
\(B=\left(\frac{7}{12}-\frac{5}{12}\right)+\left(\frac{5}{6}+\frac{1}{4}-\frac{3}{7}\right)\)
\(B=\left(\frac{1}{6}+\frac{5}{6}\right)+\frac{1}{4}-\frac{3}{7}\)
\(B=\frac{5}{4}-\frac{3}{7}\)
\(B=\frac{23}{28}\)
b)
\(x=A-B\)
\(x=\frac{1}{3}-\frac{23}{28}\)
\(x=\frac{-41}{84}\)
a) \(16^7-2^{24}\)
\(=268435456-16777216\)
\(=251658240\)
Mà \(251658240\)chia hết cho 15
\(\Rightarrow16^7-2^{24}\)chia hết cho 15
b) \(7^{80}+7^{85}-7^{84}\)
\(=7^{84}\left(7^2+7-1\right)\)
\(=7^{84}\left(49+7-1\right)\)
\(=7^{84}\left(56-1\right)\)
\(=7^{84}.55\)
Mà 55 luôn luôn chia hết cho 55
\(\Rightarrow7^{80}+7^{85}-7^{84}\)chia hết cho 55
c) \(16^5+2^{15}\)
\(16^5=2^{20}\)
\(\Rightarrow16^5+2^{15}=2^{20}+2^{15}\)
\(=2^{15}.2^5+2^{15}\)
\(=2^{15}\left(2^5+1\right)\)
\(=2^{15}.33\)
Mà 33 luôn luôn chia hết cho 33
\(\Rightarrow16^5+2^{15}\)chia hết cho 33
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