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\(\dfrac{1.3.5+2.6.10+4.12.20+7.21.35}{1.5.7+2.10.14+4.20.28+7.35.49}\)
\(=\dfrac{1.3.5+2^3.1.3.5+2^6.1.3.5+7^3.1.3.5}{1.5.7+2^3.1.5.7+2^6.1.5.7+7^3.1.5.7}\)
\(=\dfrac{1.3.5\left(1+2^3+2^6+7^3\right)}{1.5.7\left(1+2^3+2^6+7^3\right)}\)
\(=\dfrac{1.3.5}{1.5.7}\)
\(=\dfrac{3}{7}\)
Ta có : \(\dfrac{1.3.5+2.6.10+4.12.20 +7.21.35 }{1.5.7+2.10.14+4.20.28+7.35.49}\)
\(=\dfrac{1.3.5+1.2.3.2.5.2+1.4.3.4.5.4+1.7.3.7.5.7}{1.5.7+1.2.5.2.7.2+1.4.5.4.7.4+1.7.5.7.7.7}\)
\(=\dfrac{1.\left(1.3.5\right)+2.\left(1.3.5\right)+4.\left(1.3.5\right)+7.\left(1.3.5\right)}{1.\left(1.5.7\right)+2.\left(1.5.7\right)+4.\left(1.5.7\right)+7.\left(1.5.7\right)}\)
\(=\dfrac{1.3.5.\left(1+2+4+7\right)}{1.5.7.\left(1+2+4+7\right)}\)
\(=\dfrac{3}{7}\)
\(\frac{1.2.3+2.4,6+4.8.12+7.14.21}{1.3.5+2.6.10+4.12.20+7.21.35}\)
\(=\frac{1\left(1.2.3\right)+2\left(1.2.3\right)+4\left(1.2.3\right)+7\left(1.2.3\right)}{1\left(1.3.5\right)+2\left(1.3.5\right)+4\left(1.2.3\right)+7\left(1.2.3\right)}\)
\(=\frac{6\left(1+2+4+7\right)}{15\left(1+2+4+7\right)}=\frac{6}{15}=\frac{3}{5}\)
A=11.300+12.301+13.302+...+1101.400�=11.300+12.301+13.302+...+1101.400
A=1299.(11−1300+12−1301+13−13012+...+1101−1400)�=1299.(11−1300+12−1301+13−13012+...+1101−1400)
A=1299.(11−1400)�=1299.(11−1400)
A=1299.399400�=1299.399400
A=399119600�=399119600
B=11.102+12.103+13.104+...+1299.400�=11.102+12.103+13.104+...+1299.400
B=1101.(11−1102+12−1103+....+1299−1400)�=1101.(11−1102+12−1103+....+1299−1400)
B=1101.(11−1400)�=1101.(11−1400)
B=1101.399400�=1101.399400
B=39940400�=39940400
⇒AB=39911960039940400=101299
(chỉnh đề)
A=\(-1+2-3-4-5+6-7-8-9+...-2021-2022+2023-2024\)
=\(\left(-1-2024\right)+\left(2+2023\right)+\left(-3-2022\right)+\left(-4-2021\right)+\left(-5-2020\right)+\left(6+2019\right)-\left(-7-2018\right)+\left(-8-2017\right)+\left(-9-2016\right)+...+\left(1010+1015\right)+\left(-1011-1014\right)+\left(-1012-1013\right)\)=\(-2025+2025-2025-2025-2025+2025-2025-2025-2025+...+2025-2025-2025\)=253.2025-1771.2025=-3 073 950.
B=\(1.3.5+3.5.7+5.7.9+7.9.11+...+99.101.103\)
8B=\(1.3.5.8+3.5.7.8+5.7.9.8+7.9.11.8+...+99.101.103.8\)
8B=\(1.3.5.\left[7-\left(-1\right)\right]+3.5.7.\left(9-1\right)+5.7.9.\left(11-3\right)+7.9.11.\left(13-5\right)+...+99.101.103.\left(105-97\right)\)8B=\(3.5+3.5.7+3.5.7.9-3.5.7+5.7.9.11-3.5.7.9+7.9.11.13-5.7.9.11+...+99.101.103.105-97.99.101.103\)
B=\(\dfrac{3.5+99.101.103.105}{8}=13517400\)
Xét số hạng tổng quát:
1 + 1/[k.(k + 2)] = [k.(k + 2) + 1]/[k.(k + 2)] = (k + 1)²/[k.(k + 1)], với k nguyên dương.
Cho k chạy từ 1 đến 99, ta có:
• 1 + 1/1.3 = 2²/(1.3).
• 1 + 1/2.4 = 3²/(2.4).
• 1 + 1/3.5 = 4²/(3.5).
.......................
• 1 + 1/97.99 = 98²/(97.99).
• 1 + 1/98.100 = 99²/(98.100).
• 1 + 1/99.101 = 100²/(99.101).
Nhân vế với vế các đẳng thức trên, ta được:
(1 + 1/1.3).(1 + 1/2.4)(1 + 1/3.5)....(1 + 1/99.101)
= [2².3².....100²]/[1.2.3².4²......99².100...
= (2².100²)/(2.100.101)
= 2.100/101
= 200/101.
còn N thì chịu
M=(4/1.3.9/2.4.16/3.5...10000/99.101
M=2.2/1.3.3.3/2.4.4.4/3.5...100.100/99.101
M=2.3.4.5...100/1.2.3...99.3.4.5...100/2.3.4.5...101
M=100.2/101=200/101
Cau N sai de rui ban a, o mau so phai la 1.5.7+2.10.14+4.20.28+7.35.49 moi lam dc.