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19 tháng 3 2021
a) Ta có: \(1+\left(-2\right)+3+\left(-4\right)+...+2021\)
\(=\left(1-2\right)+\left(3-4\right)+...+\left(2019-2020\right)+2021\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+2021\)
\(=-1010+2021=1011\)
19 tháng 3 2021
1+(-2)+3+(-4)+....+2021
=(1+3+5+.....+2021)-(2+4+6+....+2020)
=\(\dfrac{\left(1+2021\right)\cdot2021}{2}-\dfrac{\left(2+2020\right)\cdot2020}{2}=1010\cdot2021-1010\cdot2020\\ =1010\cdot\left(2021-2020\right)=1010\cdot1\\ =1010\)
10 tháng 5 2021
Giải:
a) \(75\%+1,2-2+\dfrac{1}{5}+2018^0\)
=\(\dfrac{3}{4}+\dfrac{6}{5}-2+\dfrac{1}{5}+1\)
=\(\left(\dfrac{6}{5}+\dfrac{1}{5}\right)+\left(\dfrac{3}{4}-2+1\right)\)
=\(\dfrac{7}{5}+\dfrac{-1}{4}\)
=\(\dfrac{23}{20}\)
b) \(\left(\dfrac{-4}{3}+0,75\right):\dfrac{2017}{2018}+\left(1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\)
=\(\left(\dfrac{-4}{3}+0,75+1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\)
=\(\left[\left(\dfrac{-4}{3}+1+\dfrac{1}{3}\right)+\left(0,75-75\%\right)\right]:\dfrac{2017}{2018}\)
=\(\left[0+0\right]:\dfrac{2017}{2018}\)
=0\(:\dfrac{2017}{2018}\)
=0
10 tháng 5 2021
c)\(\left(2018-\dfrac{1}{3}-\dfrac{2}{4}-\dfrac{3}{5}-\dfrac{4}{6}-...-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\)
=\(\left(1-\dfrac{1}{3}-1-\dfrac{2}{4}-1-\dfrac{3}{5}-1-\dfrac{4}{6}-...-1-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\)
=\(\left(\dfrac{2}{3}-\dfrac{2}{4}-\dfrac{2}{5}-\dfrac{2}{6}-...-\dfrac{2}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left[2.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[\dfrac{5}{5}.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[5.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(10.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =-10
H
25 tháng 6 2023
\(a,\dfrac{1}{3}-\left(-1\dfrac{2}{5}\right)+\left(-3\dfrac{1}{4}\right)\\ =\dfrac{1}{3}-\left(-\dfrac{7}{5}\right)-\dfrac{13}{4}\\ =\dfrac{1}{3}+\dfrac{7}{5}-\dfrac{13}{4}\\ =\dfrac{20}{3\times20}+\dfrac{7\times12}{5\times12}-\dfrac{13\times15}{4\times15}\\ =\dfrac{20+84-195}{60}\\ =\dfrac{-91}{60}\)
\(b,\dfrac{5}{4}-\left(-3\dfrac{1}{2}\right)-\dfrac{7}{10}\\ =\dfrac{5}{4}+\dfrac{7}{2}-\dfrac{7}{10}\\ =\dfrac{5\times5}{4\times5}+\dfrac{7\times10}{2\times10}-\dfrac{7\times2}{10\times2}\\ =\dfrac{25+70-14}{20}\\ =\dfrac{81}{20}\)
\(c,\dfrac{3}{2}-\left[\left(-\dfrac{4}{7}-\left(\dfrac{1}{2}+\dfrac{5}{8}\right)\right)\right]\\ =\dfrac{3}{2}-\left[-\dfrac{4}{7}-\left(\dfrac{4}{8}+\dfrac{5}{8}\right)\right]\\ =\dfrac{3}{2}-\left(-\dfrac{4}{7}-\dfrac{9}{8}\right)\\ =\dfrac{3}{2}+\dfrac{4}{7}+\dfrac{9}{8}\\ =\dfrac{3\times28}{2\times28}+\dfrac{4\times8}{8\times7}+\dfrac{9\times7}{8\times7}\\ =\dfrac{84+32+63}{56}\\ =\dfrac{179}{56}\)
25 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`a,`
\(\dfrac{1}{3}-\left(-1\dfrac{2}{5}\right)+\left(-3\dfrac{1}{4}\right)\)
`= 1/3+7/5 - 13/4`
`= 26/15 - 13/4`
`= -91/60`
`b,`
\(\dfrac{5}{4}-\left(-3\dfrac{1}{2}\right)-\dfrac{7}{10}\)
`= 5/4+7/2 - 7/10`
`= 1,25 + 3,5 - 0,7`
`= 4,75 - 0,7`
`= 4,05`
`c,`
\(\dfrac{3}{2}-\left[\left(-\dfrac{4}{7}\right)-\left(\dfrac{1}{2}+\dfrac{5}{8}\right)\right]\)
`= 3/2 - [(-4/7) - 9/8]`
`= 3/2 - (-95/56)`
`= 179/56`
HC
3
31 tháng 3 2017
Đáp án của tớ là:
\(\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2003}=\)\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}\right)-\)\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1001}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}\right)-\)\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)-\)\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)=\)\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-...-\frac{1}{2002}\)\(-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-...-\frac{1}{2002}\)
Vậy:\(1+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}=\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2003}\)
6 tháng 3 2015
xin chòa hôm nay mình sẽ giúp bạn lam bài toán này
ta có
1/1002+1/1003+....+1/2003=(1+1/2+1/3+.....+1/2003)-(1+1/2+1/3+....+1/1001)
1/1002+1/1003+....+1/2003=(1+1/2+1/3+.....+1/2003)-(1/2+1/4+1/6+....+1/2002)-(1/2+1/4+1/6+......+1/2002)
1/1002+1/1003+.....+1/2003=1+1/2+1/3+....+1/2003-1/2+1/4+1/6+....+1/2002-1/2-1/4-1/6-....-1/2002
Vậy1/1002+1/1002+.....+1/2003=1-1/2+1/3-1/4+....-2/2002-1/2003
29 tháng 7 2023
a: =1/6+14/6-3/6=12/6=2
b: =-13/8+5/4:(-5/4)
=-13/8-1=-21/8
c: =-3/8(2/5+14/5)
=-3/8*16/5
=-6/5
d: =5/34(1/4+11/9-2/9+29/4)
=5/34*(15/2+1)
=5/34*17/2
=5/4
18 tháng 1 2022
Bài 1:
a: =25+75=100
b: =60-17-43+12=12
c: =-2-18=-20
d: =-3+36-17=36-20=16
Bài 2:
a: =-102
b: =-1000
c: =12x15=180
d: =21x(-10)=-210
23 tháng 3 2021
A=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A=1-1/100 A=99/100 B= (1/5.6+1/6/7+...+1/101.102).3 B=(1/5-1/6+1/6-1/7+...+1/101-1/102).3 B=(1/5-1/102).3 B=97/170
23 tháng 3 2021
1) Tính
a) Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2003}-1\right)\)
\(\Rightarrow A=\frac{-1}{2}.-\frac{2}{3}...-\frac{2002}{2003}\)
\(\Rightarrow A=\frac{1.2.3...2002}{2.3...2003}\)
\(\Rightarrow A=\frac{1}{2003}\)
A=(1/2 - 1).(1/3 - 1). ... .(1/2003 - 1)
=>A=(-1/2).(-2/3). ... .(-2002/2003)
=> A=1.2.3....2002/2.3....2003
=> A=1/2003