\(A=\frac{-2}{9}+\frac{-3}{4}+\frac{3}{5}+\frac{1}{15}+\frac{1}{57}+\frac{1}{3}+\frac{-1}{36}\)

\(A=\left(\frac{-2}{9}+\frac{-3}{4}+\frac{1}{3}+\frac{-1}{36}\right)+\left(\frac{3}{5}+\frac{1}{15}\right)+\frac{1}{57}\)

\(A=\left(\frac{-8}{36}+\frac{-27}{36}+\frac{12}{36}+\frac{-1}{36}\right)+\left(\frac{9}{15}+\frac{1}{15}\right)+\frac{1}{57}\)

\(A=\frac{-2}{3}+\frac{2}{3}+\frac{1}{57}\)

\(A=\frac{-38}{57}+\frac{38}{57}+\frac{1}{57}\)

\(A=\frac{1}{57}\)

A=1/57

B=1/41

100%

a. \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)

= \(\dfrac{-4}{6}+\dfrac{-2}{10}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)

= \(\dfrac{-3}{2}+\dfrac{1}{2}+\dfrac{3}{4}\)

= (-1) + \(\dfrac{3}{4}\)

= \(\dfrac{-4}{4}+\dfrac{3}{4}\)

= \(\dfrac{-1}{4}\)

b; 0,5  + \(\dfrac{1}{3}\) + 0,4 + \(\dfrac{5}{7}\) + \(\dfrac{1}{6}\) - \(\dfrac{4}{35}\)

= (\(\dfrac{1}{3}\)+ \(\dfrac{1}{6}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{5}{7}\)- \(\dfrac{4}{35}\)+ \(\dfrac{2}{5}\))

= ( \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{3}{5}\) + \(\dfrac{2}{5}\))

= 1 + 1 

= 2

\(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{-5}{7}+\dfrac{1}{6}-\dfrac{3}{35}+\dfrac{1}{3}-\dfrac{-1}{41}\)

\(=\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(-\dfrac{1}{5}-\dfrac{5}{7}-\dfrac{3}{35}\right)+\dfrac{1}{41}\)

\(=\dfrac{3+2+1}{6}+\dfrac{-7-25-3}{35}+\dfrac{1}{41}\)

\(=\dfrac{6}{6}+\dfrac{-35}{35}+\dfrac{1}{41}=\dfrac{1}{41}\)

Ngu

2: 

a: =>x=1/2+1/6=3/6+1/6=4/6=2/3

b: =>x+3,5=7,8-6,3=1,5

=>x=1,5-3,5=-2

c: =>(35-x)/-105=1/5

=>35-x=-21

=>x=56

1

c hình như tính bình thường thôi:v

d

= 1,75 : 5 + 2,5 . (16 – 4 . 4,1)

= 1,75 : 5 + 2,5 . (16 – 16,4)

= 1,75 : 5 + 2,5 . (−0,4)

= 0,35 − 1

= −0,65.

2

a

\(\Rightarrow x=\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{3}{6}+\dfrac{1}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)

b

\(\Rightarrow x+\dfrac{35}{10}=\dfrac{78}{10}-\dfrac{63}{5}:2\\ \Rightarrow x+\dfrac{35}{10}=\dfrac{78}{10}-\dfrac{63}{10}=\dfrac{15}{10}\\ \Rightarrow x=\dfrac{15}{10}-\dfrac{35}{10}=-2\)

c không rõ đề:v

a: Ta có: \(A=\frac{-2}{9}+\frac{-3}{4}+\frac35+\frac{1}{15}+\frac{1}{57}+\frac13+\frac{-1}{36}\)

\(=\left(-\frac29-\frac34-\frac{1}{36}\right)+\left(\frac35+\frac{1}{15}+\frac13\right)+\frac{1}{57}\)

\(=\left(-\frac{8}{36}-\frac{27}{36}-\frac{1}{36}\right)+\left(\frac{9}{15}+\frac{1}{15}+\frac{5}{15}\right)+\frac{1}{57}\)

\(=-1+1+\frac{1}{57}=\frac{1}{57}\)

b: \(B=\frac12+\frac{-1}{5}+\frac{-5}{7}+\frac16+\frac{-3}{35}+\frac13+\frac{1}{41}\)

\(=\left(\frac12+\frac13+\frac16\right)-\left(\frac15+\frac57+\frac{3}{35}\right)+\frac{1}{41}\)

\(=\left(\frac36+\frac26+\frac16\right)-\left(\frac{7}{35}+\frac{25}{35}+\frac{3}{35}\right)+\frac{1}{41}\)

\(=1-1+\frac{1}{41}=\frac{1}{41}\)

c: \(C=\frac{-1}{2}+\frac35+\frac{-1}{9}+\frac{1}{127}+\frac{-7}{18}+\frac{4}{35}+\frac27\)

\(=\left(-\frac12-\frac19-\frac{7}{18}\right)+\left(\frac35+\frac27+\frac{4}{35}\right)+\frac{1}{127}\)

\(=\left(-\frac{9}{18}-\frac{2}{18}-\frac{7}{18}\right)+\left(\frac{21}{35}+\frac{10}{35}+\frac{4}{35}\right)+\frac{1}{127}\)

\(=-1+1+\frac{1}{127}=\frac{1}{127}\)

a; - \(\dfrac{2}{3}\) + \(\dfrac{3}{4}\) - (- \(\dfrac{1}{6}\)) + (- \(\dfrac{2}{5}\))

= - \(\dfrac{2}{3}\) + \(\dfrac{3}{4}\) + \(\dfrac{1}{6}\)  - \(\dfrac{2}{5}\)

=  \(-\dfrac{40}{60}\) + \(\dfrac{45}{60}\) + \(\dfrac{10}{60}\) - \(\dfrac{24}{60}\)

=      \(\dfrac{5}{60}\) + \(\dfrac{10}{60}\) - \(\dfrac{24}{60}\)

  =      \(\dfrac{15}{60}\) - \(\dfrac{24}{60}\)

 =  - \(\dfrac{3}{20}\) 

b; (- \(\dfrac{2}{3}\)) + (- \(\dfrac{1}{5}\)) + \(\dfrac{3}{4}\) - \(\dfrac{5}{6}\) - \(\dfrac{-7}{10}\)

 = - \(\dfrac{2}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{3}{4}\) - \(\dfrac{5}{6}\) + \(\dfrac{7}{10}\)

= - \(\dfrac{40}{60}\) - \(\dfrac{12}{60}\) + \(\dfrac{45}{60}\) - \(\dfrac{50}{60}\) + \(\dfrac{42}{60}\)

= - \(\dfrac{52}{60}\) + \(\dfrac{45}{60}\) - \(\dfrac{50}{60}\) + \(\dfrac{42}{60}\)

= - \(\dfrac{7}{60}\) - \(\dfrac{50}{60}\) + \(\dfrac{42}{60}\)

= - \(\dfrac{57}{60}\) + \(\dfrac{42}{60}\)

= - \(\dfrac{1}{4}\)