Lời giải:

a. $(-2018)+2018=2018-2018=0$

b) $57+(-93)=-(93-57)=-36$

c) $(-38)+46=46-38=8$

Tính: 

a) 

b) 

c) 

`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

Cảm ơn bạn 

a: \(75\%+1,2-2+\frac15+2018^0\)

=0,75+1,2-2+0,2+1

=0,75+1,4-1

=0,75+0,4

=1,15

b: \(\left(-\frac43+0,75\right):\frac{2017}{2018}+\left(1\frac13-75\%\right):\frac{2017}{2018}\)

\(=\left(-\frac43+0,75+\frac43-0,75\right):\frac{2017}{2018}\)

=0

c: \(\left(2018-\frac13-\frac24-\cdots-\frac{2018}{2020}\right):\left(\frac{1}{15}+\frac{1}{20}+\cdots+\frac{1}{10100}\right)\)

\(=\left(1-\frac13+1-\frac24+\cdots+1-\frac{2018}{2020}\right):\left\lbrack\frac15\left(\frac13+\frac14+\cdots+\frac{1}{2020}\right)\right\rbrack\)

\(=\frac{\left(\frac23+\frac24+\cdots+\frac{2}{2020}\right)}{\frac15\left\lbrack\frac13+\frac14+\cdots+\frac{1}{2020}\right\rbrack}=2:\frac15=10\)

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)= (2018 . (157-99-1)):2018-57

   =2018.57:2018-57

   =????