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

\(-2\dfrac{3}{4}.\left(-0,4\right)+1\dfrac{3}{5}.2,75-1,2:\dfrac{4}{11}\)

\(=\dfrac{11}{4}.\dfrac{-2}{5}+\dfrac{8}{5}.\dfrac{11}{4}-\dfrac{6}{5}:\dfrac{4}{11}\)

\(=\dfrac{11}{4}.\dfrac{-2}{5}+\dfrac{8}{5}.\dfrac{11}{4}-\dfrac{6}{5}.\dfrac{11}{4}\)

\(=\dfrac{11}{4}\left(\dfrac{-2}{5}+\dfrac{8}{5}-\dfrac{6}{5}\right)\)

\(=\dfrac{11}{4}.0\)

\(=0\)

\(-2\dfrac{3}{4}.\left(-0,4\right)+1\dfrac{2}{5}.2,75-\left(-1,2\right):\dfrac{4}{11}\)

\(=\dfrac{11}{4}.\dfrac{2}{5}+\dfrac{7}{5}.\dfrac{11}{4}+\dfrac{6}{5}.\dfrac{11}{4}\)

\(=\dfrac{11}{4}.\left(\dfrac{2}{5}+\dfrac{7}{5}+\dfrac{6}{5}\right)\)

\(=\dfrac{11}{4}.\dfrac{15}{5}\)

\(=\dfrac{11}{4}.3\)

\(=\dfrac{33}{4}\)

a) Ta có: \(-3\dfrac{1}{4}\cdot x-75\%+\dfrac{3x}{2}=-1.2:\dfrac{-9}{10}-1\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{-13x}{4}-\dfrac{3}{4}+\dfrac{3x}{2}=\dfrac{-6}{5}\cdot\dfrac{10}{-9}-\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{-13x-3+6x}{4}=\dfrac{4}{3}-\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{-7x-3}{4}=\dfrac{1}{12}\)

\(\Leftrightarrow-7x-3=\dfrac{1}{3}\)

\(\Leftrightarrow-7x=\dfrac{10}{3}\)

hay \(x=-\dfrac{10}{21}\)

b) Ta có: \(\dfrac{5}{3}+\dfrac{5}{15}+\dfrac{5}{35}+...+\dfrac{5}{x\left(x+2\right)}=2\dfrac{8}{17}\)

\(\Leftrightarrow\dfrac{5}{2}\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{x\left(x+2\right)}\right)=2\dfrac{8}{17}\)

\(\Leftrightarrow\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=2+\dfrac{8}{17}\)

\(\Leftrightarrow\left(1-\dfrac{1}{x+2}\right)=\dfrac{42}{17}:\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{x+1}{x+2}=\dfrac{42}{17}\cdot\dfrac{2}{5}=\dfrac{84}{85}\)

\(\Leftrightarrow85x+85=84x+168\)

\(\Leftrightarrow x=83\)

a.\(\dfrac{27}{8}\)

b.\(\dfrac{37}{40}\)
c.\(\dfrac{5}{2}\)

d.\(\dfrac{7}{3}\)

e.5

g.\(\dfrac{53}{16}\)

Bài 1 :

a) \(\dfrac{3}{2}+\dfrac{5}{4}+\dfrac{5}{8}=\dfrac{12}{8}+\dfrac{10}{8}+\dfrac{5}{8}=\dfrac{12+10+5}{8}=\dfrac{27}{8}\)

b) \(\dfrac{4}{5}-\dfrac{3}{8}+\dfrac{2}{4}=\dfrac{32}{40}-\dfrac{15}{40}+\dfrac{20}{40}=\dfrac{32-15+20}{40}=\dfrac{37}{40}\)

c) \(3+\dfrac{6}{8}-\dfrac{5}{4}=\dfrac{3}{1}+\dfrac{6}{8}-\dfrac{5}{4}=\dfrac{24}{8}+\dfrac{6}{8}-\dfrac{10}{8}=\dfrac{20}{8}=\dfrac{5}{2}\)

d) \(\dfrac{5}{6}-\dfrac{1}{2}+2=\dfrac{5}{6}-\dfrac{1}{2}+\dfrac{2}{1}=\dfrac{5}{6}-\dfrac{3}{6}+\dfrac{12}{6}=\dfrac{14}{6}=\dfrac{7}{3}\)

e) \(\dfrac{3}{5}+\dfrac{6}{11}+\dfrac{7}{13}+\dfrac{2}{5}+\dfrac{16}{11}+\dfrac{19}{13}=\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(\dfrac{6}{11}+\dfrac{16}{11}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)=1+2+2=5\)

g) \(\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{29}{32}+\dfrac{1}{4}+\dfrac{3}{21}+\dfrac{13}{32}=\dfrac{3}{4}+\dfrac{6}{7}+\dfrac{29}{32}+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{13}{32}=\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{6}{7}+\dfrac{1}{7}\right)+\left(\dfrac{29}{32}+\dfrac{13}{32}\right)=1+1+\dfrac{21}{16}=2+\dfrac{21}{16}=\dfrac{53}{16}\)

a,3/2+5/4+5/8 = 12/8 + 10/8 + 5/8 = 27/8

b,4/5-3/8+2/4 = 32/40 - 15/40 + 20/40 = 37/40

c,3+6/8-5/4= 24/8 + 6/8 - 10/8 = 20/8 = 5/2

d,5/6-1/2+2 = 5/6 - 3/6 + 12/6 = 14/6 = 7/3

e,3/5+6/11+7/13+2/5+16/11+19/13 = ( 3/5+ 2/5) + ( 6/11 + 16/11) + ( 7/13 + 19/13) = 1 + 2 + 2 = 5

g,75/100+18/21+29/32+1/4+3/21+13/32 = 3/4 + 6/7 + 29/32 + 1/4 + 1/7 + 13/32=  ( 3/4 + 1/4) + ( 29/32 + 13/32) + ( 6/7 + 1/7)= 1 + 21/16 + 1 = 53/16

a: =(5/7+2/7)+(4/3+5/3)=3+1=4

b: =(17/12+7/12)+(29/7-8/7)

=2+3=5

c: =(2/5+3/5)+(6/9+1/3)+(7/4+1/4)

=1+2+1

=4

d: =(1/3+2/3)+(13/17+4/17)+(29/11+4/11)

=1+1+3=5

\(=\dfrac{11}{4}\cdot\dfrac{2}{5}-\dfrac{8}{5}\cdot\dfrac{11}{4}+\dfrac{-6}{5}\cdot\dfrac{11}{4}\)

=11/4x(-12/5)=-132/20=-33/5