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
\(A=1.2+2.3+3.4+4.5+............+2017.2018\)
\(3A = 1.2.3 + 2.3.4 +..............+ 2017.1018.3\)
\(3A = 1.2.3 + 2.3.(4-1) + .............. + 2017.2018.(2019-2016)\)
\(3A = 1.2.3 + 2.3.4 - 1.2.3 + ............. + 2017.2018.2019 - 2016.2017.2018\)
\(3A = 2017.2018.2019\)
\(A = \frac{2017.2018.2019}{3}\)
\(B =\frac {2018^3}{3}\)
đến đây ko bt lm
A = 1.2 + 2.3 + 3.4 + 4.5 + .............. + 2017.2018
\(\Rightarrow\) 3A = 1.2.3 + 2.3.3 + 3.4.3 + 4.5.3 + .............. + 2017.2018.3
\(\Rightarrow\)3A = 1.2.3 + 2.3(4-1) + 3.4(5-2) + 4.5(6-3) + .............. + 2017.2018(2019-2016)
\(\Rightarrow\)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
\(\Rightarrow\)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 )
\(\Rightarrow\)3A = 2017.2018.2019
\(\Rightarrow\) A = \(\frac{2017.2018.2019}{3}\)
Ta có:
\(A=\frac{2017.2018.2019}{3}\) ; \(B=\frac{2018.2018.2018}{3}\)
2017 . 2019 = 2017 . ( 2018 + 1 ) = 2017 . 2018 + 2017
2018 . 2018 = 2018 . ( 2017 + 1 ) = 2018 . 2017 + 2018
Mà 2017 . 2018 + 2017 < 2018 . 2017 + 2018
\(\Rightarrow\frac{2017.2018.2019}{3}\) < \(\frac{2018.2018.2018}{3}\)
\(\Rightarrow\) A < B
Vậy A < B