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\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)
\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)
\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)
\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)
\(C=\frac{91}{9}\)
\(b)\) \(1.2.3...9-1.2.3...8-1.2.3...8^2\)
\(=\)\(1.2.3...8\left(9-1-8\right)\)
\(=\)\(1.2.3...8\left(9-9\right)\)
\(=\)\(1.2.3...8.0\)
\(=\)\(0\)
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
Câu a:
(7\(^{2017}\) - 7\(^{2015}\)) ⋮ (7\(^{2014}\).7)
= 7\(^{2015}\).(7\(^2\) - 1) ⋮ 7\(^{2015}\)
= (7\(^{2015}\) : 7\(^{2015}\)).(49 - 1)
= 1.48
= 48
Câu b:
(2\(^{13}\) + 2\(^5\)) : (2\(^{10}\) + 2\(^2\))
= 2\(^5.\)(2\(^8\) + 1):[2\(^2\).(2\(^8\) + 1)]
= (2\(^5:2^2\)).[(\(2^8+1):\left(2^8+1\right)\)]
= 2\(^3\).1
= 8
a)\(\left(10^2+11^2+12^2\right)\div\left(13^2+14^2\right)\)
\(=\left(100+121+144\right)\div\left(169+196\right)\)
\(=365\div365\)
\(=1\)
b) \(1.2.3...9-1.2.3...8-1.2.3...8^2\)
\(=1.2.3...8\left(9-1-8\right)\)
\(=1.2.3...8.0\)
\(=0\)
d) \(1152-\left(374+1152\right)+\left(-65+374\right)\)
\(=1152-374-1152-65+374\)
\(=\left(1152-1152\right)-65+\left(374-374\right)\)
\(=0-65+0\)
\(=-65\)
e) \(13-12+11+10-9+8-7-6+5-4+3+2-1\)
\(=13-\left(12-11\right)+\left(10-9\right)+\left(8-7\right)-\left(6-5\right)-\left(4-3\right)\)\(+\left(2-1\right)\)
\(=13-1+1+1-1-1+1\)
\(=13+0+0+0\)
\(=13\)
Bài 1 câu a:
1/-8 + -5/8
= -1/8 - 5/8
= -6/8
= -3/4
Câu b:
-1/21 + -1/28
= -4/84 - 3/84
= - 7/854
Câu c:
-5/9 + 8/15 + -2/11 + 4/-9 + 7/15
= (8/15+ 7/15) - (5/9 + 4/9) - 2/11
= 1 - 1 - 2/11
= 0 - 2/11
= - 2/11
Câu d:
-25/-75 + 17/34 + 121/- 132
= 1/3 + 1/2 - 11/12
= 4/12 + 6/12 - 11/12
= 10/12 - 11/12
= - 1/12
Bài 2a:
51/23 = 5151/2323 = 515151/232323
\(\frac{5151}{2323}\) = \(\frac{5151:101}{2323:101}\) = \(\frac{51}{23}\) (1)
\(\frac{515151}{232323}\) = \(\frac{515151:10101}{232323:10101}\) = \(\frac{51}{23}\) (2)
Từ (1) và (2) ta có:
\(\frac{51}{23}=\frac{5151}{2323}=\frac{515151}{232323}\) (đpcm)
sau 58 ,7,50 ,2,-60 là phép tính khác .
mấy b giúp mk nha ?