Bài 3:Tính
a,(-2018)+2018 b,57+(-93) c,(-38)+46
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Lời giải:
a. $(-2018)+2018=2018-2018=0$
b) $57+(-93)=-(93-57)=-36$
c) $(-38)+46=46-38=8$
`a, = 53 xx (12+172 -84)`
`= 53 xx 100`
`=5300`
`b,= 2018 xx (91-45-46)`
`= 2018 xx 0`
`=0`
a: =-48+27+56-48-27-36
=-96+20
=-76
b: =23-57+57-33=-10
c: =-98+12-159-12-41
=-98-200
=-298
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
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
a)36:32+23.22-33.3
= 34+25-34
=25
=32
b)38:34-95:93
= 34-92
= 34-34
= 0
c)23.15+23.35
= 23(15+35)
= 8.50
= 400
a: \(3^6:3^2+2^3\cdot2^2-3^3\cdot3\)
\(=3^4+32-3^4\)
=32
b: \(3^8:3^4-9^5:9^3\)
\(=3^4-9^2\)
=0
c: Ta có: \(2^3\cdot15+2^3\cdot35\)
\(=8\cdot50\)
=400
A = ( 157 x 2018 - 99 x 2018-2018 ) : 2018 - 57
= [2018 x ( 157 - 99 ) - 2018] : 2018 - 57
= [ 2018 x 58 - 2018 ] : 2018 - 57
= 115026 : 2018 - 57
= 57 - 57
= 0
B = ( 1 + 3 + 5 + 7 +...+ 2018 ) x ( 135135 x 137 - 135 x 137 )
( 1 + 3 + 5 + 7 +...+ 2018 ) = ( 2018 - 1 ) : 2 + 1 = 1009 số
( 1 + 3 + 5 + 7 +...+ 2018 ) = ( 2018 + 1 ) x 1009 : 2 = 1018585
Vậy B = ( 1 + 3 + 5 + 7 +...+ 2018 ) x ( 135135 x 137 - 135 x 137 )
B = 1018585 x 18495000 = số rất lớn
nếu bạn viết nhầm + thành x thì 1018585 + 18495000 = 19513588
nhớ k cho mình nhé !!!!
a=0
A=1+(-3)+5+(-7)+...+17