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\(=\left[3^{22}. 3^5+2.3^{27}\right]:3^{26}\)
\(=\left[3^{27}+2.3^{27}\right]:3^{26}\)
\(=\left[3^{27}.\left(1+2\right)\right]:3^{26}\)
\(=\left[3^{27}.3\right]:3^{26}\)
\(=3^{28}:3^{26}\)
\(=3^2\)
321 + 322 + 323 + 324 + 325 +326 + 327 + 328 + 329
= \(3^{21}.\left(1+3+3^2\right)+3^{24}.\left(1+3+3^2\right)+3^{27}.\left(1+3+3^2\right)\)
= \(3^{21}.13+3^{24}.13+3^{27}.13\)
= \(13.\left(3^{21}+3^{24}+3^{27}\right)\)
vì \(13⋮13\) nên \(13.\left(3^{21}+3^{24}+3^{27}\right)⋮13\)
vậy 321 + 322 + 323 + 324 + 325 +326 + 327 + 328 + 329 chia hết cho 13
Sửa đề: \(\frac{2\cdot4^{14}\cdot9^8-3^{17}\cdot8^9}{19\cdot2^{26}\cdot3^{13}-27^5}\)
Ta có: \(\frac{2\cdot4^{14}\cdot9^8-3^{17}\cdot8^9}{19\cdot2^{26}\cdot3^{13}-27^5\cdot2^{24}}\)
\(=\frac{2\cdot\left(2^2\right)^{14}\cdot\left(3^2\right)^8-3^{17}\cdot\left(2^3\right)^9}{19^{}\cdot2^{26}\cdot3^{13}-\left(3^3\right)^5\cdot2^{24}}\) \(=\frac{2^{29}\cdot3^{16}-3^{17}\cdot2^{27}}{19\cdot2^{26}\cdot3^{13}-3^{15}\cdot2^{24}}=\frac{2^{27}\cdot3^{16}\left(2^2-3\right)}{2^{24}\cdot3^{13}\left(19\cdot2^2-3^2\right)}\)
\(=\frac{2^3\cdot3^3}{19\cdot4-9}=\frac{8\cdot9}{76-9}=\frac{72}{67}\)
Sửa đề: \(\frac{2\cdot4^{14}\cdot9^8-3^{17}\cdot8^9}{19\cdot2^{26}\cdot3^{13}-27^5}\)
Ta có: \(\frac{2\cdot4^{14}\cdot9^8-3^{17}\cdot8^9}{19\cdot2^{26}\cdot3^{13}-27^5\cdot2^{24}}\)
\(=\frac{2\cdot2^{28}\cdot3^{16}-3^{17}\cdot2^{27}}{19\cdot2^{26}\cdot3^{13}-3^{15}\cdot2^{24}}\)
\(=\frac{2^{27}\cdot3^{16}\left(2^2-3\right)}{3^{13}\cdot2^{24}\cdot\left(19\cdot2^2-3^2\right)}=\frac{2^3\cdot3\cdot1}{76-9}=\frac{24\cdot3}{67}=\frac{72}{67}\)
\(\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{22.3^{28}-3^{28}.2}{2^3.3^{28}}=\frac{3^{28}.20}{3^{28}.4}=\frac{20}{4}=5\)
1,2 dễ ko làm
3,
S = 1 + 2 + 22 + 23 + ... + 29
2S = 2 + 22 + 23 + 24 + ... + 210
2S - S = ( 2 + 22 + 23 + 24 + ... + 210 ) - ( 1 + 2 + 22 + 23 + ... + 29 )
S = 210 - 1
Mà 5 . 28 = ( 1 + 22 ) . 28 = 28 + 210 > 210 > 210 - 1
Vậy S < 5 . 28
P = 1 + 3 + 32 + 33 + ... + 320
3P = 3 + 32 + 33 + 34 + ... + 321
3P - P = ( 3 + 32 + 33 + 34 + ... + 321 ) - ( 1 + 3 + 32 + 33 + ... + 320 )
2P = 321 - 1
P = ( 321 - 1 ) : 2 < 321
Vậy P < 321
\(\left[3^{22}\cdot\left(3^8:3^3\right)+2\cdot3^{27}\right]:3^{26}\\ =\left(3^{22}\cdot3^5+2\cdot3^{27}\right):3^{26}\\ =3^{27}\left(1+2\right):3^{26}\\ =3^{27}\cdot3:3^{26}\\ =3^2=9\)
\(\left[3^{22}.\left(3^8:3^3\right)+2.3^{27}\right]\):\(3^{26}\)
=\(\left(3^{22}.3^5+2.3^{27}\right):3^{26}\)
=(\(3^{27}\)+2.\(3^{27}\)):\(3^{26}\)
=\(3^{27}\).(1+2):\(3^{26}\)
=\(3^{28}:3^{26}=3^2\)=9