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a) 1+2+3+...+2018
=[(2018-1):1+1].(2018+1):2
=2018.2019:2
=2037171
b) 4+7+10+13+...+2017
=[(2017-4):3+1].(2017+4):2
=672.2021:2
=679056
c) 1+2-3-4+5+6-7-8+9+10-11-12+13+14
=(1+2-3-4)+(5+6-7-8)+(9+10-11-12)+13+14
=-4+(-4)+(-4)+13+14
=-12+27
=15
d) 23 x 75 + 25 x 23 + 180
=23 x (75+25) +180
=23 x 100 +180
=2300+180
2480
chúc ban hoc tốt nha
a) Số số hạng: ( 2018 - 1 ) : 1 + 1 = 2018
Tổng: 2018 x ( 2018 + 1 ) : 2 = 2 037 171
\(\frac{2017}{2018}\)x\(\frac{7}{8}\)+\(\frac{2017}{2018}\)x\(\frac{3}{8}\)-\(\frac{2017}{2018}\)x\(\frac{1}{4}\)
= \(\frac{2017}{2018}\)x (\(\frac{7}{8}\)+\(\frac{3}{8}\)-\(\frac{1}{4}\))
= \(\frac{2017}{2018}\)x ( \(\frac{10}{8}\)- \(\frac{1}{4}\))
= \(\frac{2017}{2018}\)x ( \(\frac{10}{8}\)- \(\frac{2}{8}\))
= \(\frac{2017}{2018}\)x 1
= \(\frac{2017}{2018}\)
Chúc em học tốt nhé :>
=2017/2018*(7/8+3/8)-2017*1/4
=2017/2018*5/4+2017*-1/4
=2017/2018*(5/4-1/4)
=2017/2018*1
=2017/2018
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}>1\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{2017.2018}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2017}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}\)
\(=\frac{504}{1009}\)
a, Bài giải
Ta có : 2018 : 13 = 155 (dư 3)
Vậy số cần tìm là :
13 - 3 = 10
Đáp số : 10
b, \(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{2017\times2018}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}\)
\(=\frac{504}{1009}\)
a) ( 1/2-1/3-1/6).(1/2+2/3+3/4+...+2017/2018) + 3/4.x = 9/10
0.(1/2+2/3+3/4+...+2017/2018) + 3/4.x = 9/10
0+3/4.x = 9/10
3/4.x = 9/10
x = 9/10: 3/4
x = 6/5
b) x + ( 3/1.3+3/3.5+...+3/13.15) = 11/5
x + 3/2. ( 1-1/3 + 1/3 - 1/5 + ...+ 1/13 - 1/15) = 11/5
x + 3/2. ( 1-1/15) = 11/5
x + 3/2.14/15 = 11/5
x + 7/5 = 11/5
x = 11/5 - 7/5
x = 4/5
= (1-1/2018)-(1+1/2018)-2020/2019
= 1-1/2018-1-1/2018-2020/2019
= -2/2018-2020/2019
vậy thôi
=(1-1/2018)-(1+1/2018)-2020/2019
=1-1/2018-1-1/2018-2020/2019
=-2/2018-2020/2019
`@` `\text {Ans}`
`\downarrow`
`a)`
`13/50 + 9% + 41/100 + 0,24`
`= 0,26 + 0,09 + 0,41 + 0,24`
`= (0,26 + 0,24) + (0,09 + 0,41)`
`= 0,5 + 0,5`
`= 1`
`b)`
`2018 \times 2020 - 1/2017 + 2018 \times 2019`
`= 2018 \times (2020 + 2019) - 1/2017`
`= 2018 \times 4039 - 1/2017`
`= 8150702`
`c)`
`1/2 + 1/6 + 1/12 + 1/20 +1/30 +1/42`
`=`\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}\)
`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{6}-\dfrac{1}{7}\)
`=`\(1-\dfrac{1}{7}\)
`= 6/7`
\(a,\dfrac{13}{50}+9\%+\dfrac{41}{100}+0,24\\ 0,26+0,09+0,41+0,24\\ =\left(0,26+0,24\right)+\left(0,09+0,41\right)\\ =0,5+0,5\\ =1\\ b,2018\times2020-\dfrac{1}{2017}+2018\times2019\\ =2018\times\left(2020+2019\right)-\dfrac{1}{2017}\\ =2018\times4039-\dfrac{1}{2017}\\ =3150702-\dfrac{1}{2017}\\ c,\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\\ =1-\dfrac{1}{2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}.........+\dfrac{1}{6}-\dfrac{1}{7}\\ =1-\dfrac{1}{7}\\ =\dfrac{6}{7}\)