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A = 1 + 2 + 3 + ... + 2018 (có 2018 số )
= (2018 + 1) . 2018 : 2 = 2037171
B = 1 + 3 + 5 + ... + 2017(có 1009 số )
= (2017 + 1) . 1009 : 2 = 1018081
C = 2 + 4 + 6 + ... + 2018 (Có 1009 số )
= (2018 + 2) x 1009 : 2 = 1019090
D = 72 . 153 + 27.153 + 153
= (72 + 27 + 1) . 153
= 100 . 153 = 15300
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}=\frac{2017}{2018}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\cdot\left(1-\frac{1}{2018}\right)\)
\(C=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2016}{2017}\cdot\frac{2017}{2018}\)
\(C=\frac{1\cdot2\cdot3\cdot...\cdot2016\cdot2017}{2\cdot3\cdot4\cdot...\cdot2017\cdot2018}\)
\(C=\frac{1}{2018}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đề bài
= A x ( 5/6 - 1/2 -1/3 )
= A x ( 5/6 - 3/6 - 2/6 )
= A x 0
= 0
Thế nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
a,0,36.350+1,2.20.3+9.4.4,5
=13.3.35+12.2.3+9.2.3.3
=3.(13.35+12.2+.9.2.3)
=3.(455+24+54)
=3.533
=1599
b,2015.2016-5/2015.2015+2010
=4062240-5+2010
=4064245
c,2/1.3+2/3.5+2/5.7+...+2/71.73
=1-1/3+1/3-1/5+1/5-1/7+...+1/71-1/73
=1-1/73
=72/73
d,(1+1/2).(1+1/3)+...+(1+1/2018)
=3/2.4/3.5/4+...+2019/2018
=2019/2
e,E=1/4.5+1/5.6+1/6.7+...+1/80.81(làm tương tự với phần d nên mình làm ngắn
=1/4-1/81
=77/324
f,F=3/2.3+3/3.4+...+3/99.100
=3.(1/2.3+1/3.4+...+1/99.100)(làm tương tự với d
=3.(1/2-1/100)
=3.49/100
=147/100
gG=5/1.4+5/4.7+...+5/61.64
3G=5.(3/1.4+3./4.7+...+3/61.64)
=5.(1-1/64)
=5.63/64
=315/64
ok nha bạn,mình giữ đúng lời hứa.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(D=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...+\left(1+\frac{1}{2018}\right)\)
\(=\frac{3}{2}.\frac{4}{3}......\frac{2018}{2017}.\frac{2019}{2018}\)
\(=\frac{3.4.5....2018.2019}{2.3.4.5....2017.2018}=\frac{2019}{2}\)
\(E=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{79.80}+\frac{1}{80.81}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{80}-\frac{1}{81}\)
\(=\frac{1}{4}-\frac{1}{81}\)
\(=\frac{77}{324}\)
\(\text{D}=\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot...\cdot\left(1+\frac{1}{2017}\right)\cdot\left(1+\frac{1}{2018}\right)\)
\(\text{D}=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{2018}{2017}\cdot\frac{2019}{2018}\)
\(\text{D}=\frac{2019}{2}\)
\(\text{E}=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{79.80}+\frac{1}{80.81}\)
\(\text{E}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{79}-\frac{1}{80}+\frac{1}{80}-\frac{1}{81}\)
\(\text{E}=\frac{1}{4}-\frac{1}{81}=\frac{81}{324}-\frac{4}{324}=\frac{77}{324}\)
\(B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2016}{2017}.\frac{2017}{2018}\)
\(\Rightarrow B=\frac{1.2.3....2016.2017}{2.3.4...2017.2018}\)
\(\Rightarrow B=\frac{1}{2018}\)
B=1/2x2/3/3/4/4/5/....../2016/2017x2017/2018
B=1/2018