\(=200-100:5+1\)

\(=200-20+1\)

=\(180+1\)

\(=181\)

181

5: =5/7(5/11+2/11-14/11)

=5/7*(-7/11)

=-5/11

6: =-5/4(2/11+9/11)+1+3/4

=7/4-5/4=1/2

7: =-5/7(3/13+10/13)+1+5/7

=1

8: =7/4(29/5-9/5)+3+2/13

=3+2/13+7/4*4

=10+2/13

=132/13

(-26) + 16 + (-34) + 26          

`=(-26+26)+[16+(-34)]`

`=0+(-18)`

`=-18`                         

\((-26) + 16 + ( - 34) + 26 \)

\(=-26+16-34+26\)

\(=(26-26)+(16-34)\)

\(=0-18\)

\(=-18\)

(-26) + 16 + (-34) + 26          

\(5:\left(-\dfrac{5}{2}\right)^2+\dfrac{2}{15}\cdot\sqrt{\dfrac{9}{4}}-\left(-2020\right)^0+\left|-0,25\right|\)

\(=5:\dfrac{25}{4}+\dfrac{2}{15}\cdot\dfrac{3}{2}-1+0,25\)

\(=5\cdot\dfrac{4}{25}+\dfrac{3}{15}-0,75\)

\(=\dfrac{4}{5}+\dfrac{1}{5}-\dfrac{3}{4}\)

\(=1-\dfrac{3}{4}\)

\(=\dfrac{1}{4}\)

\(5:\left(-\dfrac{5}{2}\right)^2+\dfrac{2}{15}\cdot\sqrt{\dfrac{9}{4}}-\left(-2020\right)^0+\left|-0,25\right|\)
\(=5:\dfrac{25}{4}+\dfrac{2}{15}\cdot\dfrac{3}{2}-1+0,25\)
\(=\dfrac{4}{5}+\dfrac{1}{5}-1+\dfrac{1}{4}\)
\(=1-1+\dfrac{1}{4}\)
\(=\dfrac{1}{4}\)

#kễnh

c: \(=\left(\dfrac{3}{5}+\dfrac{83}{200}-\dfrac{3}{200}\right)\cdot\dfrac{8}{3}\cdot\dfrac{1}{4}=\dfrac{8}{3}\cdot\dfrac{1}{4}=\dfrac{2}{3}\)

g: \(=\dfrac{1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{3}{4}\cdot\dfrac{5}{2}+\dfrac{3}{5}=\dfrac{5}{4}+\dfrac{3}{5}=\dfrac{25+12}{20}=\dfrac{37}{20}\)

C

a)-5/12-3:9/4=-5/12+(-4/3)=-7/4

c)(3/4-1/6):2/3=7/12:2/3=7/8

b)2/3-4.(1/2+3/4)=2/3-2-3=-13/3

d)5/3+-8/3:(-4/9)=5/3+6=23/3

c) = 2/3

g) = 37/20 = 1,85

n) = 3/17

e: \(=\dfrac{21-6}{8}\cdot\dfrac{16}{3}=\dfrac{15}{3}\cdot2=10\)

f: \(=\dfrac{-3\cdot3+7\cdot4}{12}:\dfrac{2\cdot2-5\cdot9}{18}\)

\(=\dfrac{19}{12}\cdot\dfrac{18}{-41}=\dfrac{-57}{82}\)

g: \(=\dfrac{9-8}{10}\cdot\dfrac{5}{2}+1=\dfrac{1}{10}\cdot\dfrac{5}{2}+1=\dfrac{1}{4}+1=\dfrac{5}{4}\)

o: \(=\dfrac{3}{4}\cdot\left[\dfrac{1}{5}-\dfrac{4}{7}-\dfrac{3}{5}\cdot\dfrac{-5}{7}\right]=\dfrac{3}{4}\cdot\left(\dfrac{1}{5}-\dfrac{4}{7}+\dfrac{3}{7}\right)\)

\(=\dfrac{3}{4}\cdot\left(\dfrac{1}{5}-\dfrac{1}{7}\right)=\dfrac{3}{4}\cdot\dfrac{2}{35}=\dfrac{3}{70}\)

p: \(=\left[\left(\dfrac{1}{9}\cdot\dfrac{27}{8}\right)\cdot3\right]\cdot\dfrac{-128}{81}=\dfrac{9}{8}\cdot\dfrac{-128}{81}=\dfrac{-16}{9}\)