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

Cách 1:
Xét số bị trừ, ta có:
(2/3 + 3/4 + 4/5 + ... + 2016/2017) x (1/2 + 2/3 + 3/4 + ... + 2015/2016)
= (2/3 + 3/4 + 4/5 + ... + 2015/2016 + 2016/2017) x (1/2 + 2/3 + 3/4 + ... + 2015/2016)
= (2/3 + 3/4 + 4/5 + ... + 2015/2016) x (1/2 + 2/3 + 3/4 + ... + 2015/2016) + 2016/2017 x (1/2 + 2/3 + 3/4 + ... + 2015/2016)
Xét số trừ, ta có: 
(1/2 + 2/3 + 3/4 + ... + 2016/2017) x (2/3 + 3/4 + 4/5 + ... + 2015/2016)
= (1/2 + 2/3 + 3/4 + ... + 2015/2016 + 2016/2017) x (2/3 + 3/4 + 4/5 + ... + 2015/2016)
= (1/2 + 2/3 + 3/4 + ... + 2015/2016) x (2/3 + 3/4 + 4/5 + ... + 2015/2016) + 2016/2017 x (2/3 +3/4 + 4/5 + ... + 2015/2016) =
Ta thấy số bị trừ và số trừ có số hạng giống nhau là:
(2/3 +3/4 + 4/5 + ... + 2015/2016) x (1/2 + 2/3 + 3/4 + ... + 2015/2016)
Nên phép trừ trên có thể viết lại:
2016/2017 x (1/2 + 2/3 + 3/4 + ... + 2015/2016) - 2016/2017 x (2/3 + 3/4 + 4/5 + ... + 2015/2016)
= 2016/2017 x [(1/2 + 2/3 + 3/4 + ... + 2015/2016) - (2/3 +3/4 + 4/5 + ... + 2015/2016)]
= 2016/2017 x 1/2
= 1008/2017

Cách 2:

tong cac so tren la:

2017x2017=3016

đáp số:3016