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19 tháng 3 2021
a) Ta có: \(1+\left(-2\right)+3+\left(-4\right)+...+2021\)
\(=\left(1-2\right)+\left(3-4\right)+...+\left(2019-2020\right)+2021\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+2021\)
\(=-1010+2021=1011\)
OD
0
OD
1
27 tháng 4 2021
-27/5 . -4/15 + -4/15 . ( 2021 - 27/5)
=-4/15.[(-27/5)+2021-27/5]
=-4/15.2012,2
= -40244/75
OD
0
OD
1
27 tháng 4 2021
-27/5 . -4/15 + -4/15 . ( 2021 - 27/5)
=-4/15.[(-27/5)+2021-27/5]
=-4/15.2012,2
= -40244/75
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`
23 tháng 6 2021
\(2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)\)
\(=\left(1-1\right)+\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+...+\left(1-\dfrac{1}{2021}\right)\)
\(=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\)
24 tháng 6 2021
Giải:
\(2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\)
Ta có:
\(2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)\)
\(=\left(1-1\right)+\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+...+\left(1-\dfrac{1}{2021}\right)\)
\(=0+\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{2020}{2021}\)
\(=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{2020}{2021}\)
Mà \(\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{2020}{2021}=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\)
\(\Rightarrow2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\left(đpcm\right)\)
1+(-2)+3+(-4)+....+2021
=(1+3+5+.....+2021)-(2+4+6+....+2020)
=\(\dfrac{\left(1+2021\right)\cdot2021}{2}-\dfrac{\left(2+2020\right)\cdot2020}{2}=1010\cdot2021-1010\cdot2020\\ =1010\cdot\left(2021-2020\right)=1010\cdot1\\ =1010\)
1 + (-2) + 3 + (-4) +......+ 2021
= ( 1 - 2 ) + ( 3 - 4 ) +....+ ( 2019 - 2020 ) + 2021
= (-1) + (-1) +.....+ (-1) + 2021
Từ 1 đến 2020 có số số hạng là :
( 2020 - 1 ) : 1 + 1 = 2020 ( số hạng )
⇒ 1 + 2021 = 2022