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cho bài kham khảo nè :
A=1.2+2.3+3.4+4.5+...+2017.2018
=> 3A=1.2.3+2.3.3+3.4.3+4.5.3+...+2017.2018.3
3A=1.2.3+2.3(4-1)+3.4(5-2)+4.5(6-3)+...+2017.2018.(2019-2016)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+2017.2018.2019-2016.2017.2018
3A=(1.2.3+2.3.4+3.4.5+4.5.6+...+2017.2018.2019)-(1.2.3+2.3.4+3.4.5+...+2016.2017.2018)
=> 3A=2017.2018.2019 => \(A=\frac{2017.2018.2019}{3};B=\frac{2018^3}{3}=\frac{2018.2018.2018}{3}\)
Ta có: 2017.2019=2017(2018-1)=2017.2018+2017<2017.2018+2018=2018(2017+1)=2018.2018
=> 2017.2018.2019<2018.2018.2018
=> A<B
thank nha
A=1.2+2.3+3.4+...+2017.2018
3A=1.2.3+2.3.3+3.4.3+...+2017.2018.3
3A=1.2.3+2.3.(4−1)+3.4.(5−2)+...+2017.2018.(2019−2016)
3A=1.2.3+2.3.4−1.2.3+3.4.5−2.3.4+...+2017.2018.2019−2016.2017.2018
⇒3A=2017.2018.2019
⇒A=2017.2018.20193
A=2017.2018.20193;B=201833=2018.2018.20183
A=2739315938;B=2739316611
⇒A<B
A=1.2+2.3+3.4+4.5+...+2017.2018
=> 3A=1.2.3+2.3.3+3.4.3+4.5.3+...+2017.2018.3
3A=1.2.3+2.3(4-1)+3.4(5-2)+4.5(6-3)+...+2017.2018.(2019-2016)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+2017.2018.2019-2016.2017.2018
3A=(1.2.3+2.3.4+3.4.5+4.5.6+...+2017.2018.2019)-(1.2.3+2.3.4+3.4.5+...+2016.2017.2018)
=> 3A=2017.2018.2019 => \(A=\frac{2017.2018.2019}{3}\); \(B=\frac{2018^3}{3}=\frac{2018.2018.2018}{3}\)
Ta có: 2017.2019=2017(2018-1)=2017.2018+2017<2017.2018+2018=2018(2017+1)=2018.2018
=> 2017.2018.2019<2018.2018.2018
=> A<B
Bui The Hao lam dung roi
mk cung dang can bai nay
Thanks vi da dang honganh
a, so sánh (a+1).(a+2).(a+3)-a.(a+1).(a=2) và 3. (a+1).(a+2)
b/ tính P = 1.2+2.3+3.4+...+2004.2005
Chị dùg cách tính tổng đi
1. Tìm dãy cách đều bao nhiêu
2. Từ công thức tính tổng rồi suy ra
B = 1.(1+1) + 2.(2+1) + 3.(3+1) +...+ 14.(14 +1) = (12 + 22 + 32 +...+142 ) + (1+2+3+...+14)
=> B - A = (1+2+3+....+14) - 152 = [(1+14).14 : 2] - 225 = -120