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

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

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

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

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

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

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

c; \(\dfrac{1}{3}\) + \(\dfrac{-1}{5}\) - \(\dfrac{-1}{4}\) - \(\dfrac{7}{9}\)- \(\dfrac{-5}{12}\) + \(\dfrac{-1}{45}\) + \(\dfrac{1}{41}\)

  = (\(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) + \(\dfrac{5}{12}\)) - (\(\dfrac{1}{5}\) + \(\dfrac{7}{9}\) + \(\dfrac{1}{45}\)) + \(\dfrac{1}{41}\)

= (\(\dfrac{4}{12}\) + \(\dfrac{3}{12}\) + \(\dfrac{5}{12}\)) - (\(\dfrac{9}{45}+\dfrac{35}{45}\) + \(\dfrac{1}{45}\)) + \(\dfrac{1}{41}\)

= (\(\dfrac{7}{12}\) + \(\dfrac{5}{12}\)) - (\(\dfrac{44}{45}\) + \(\dfrac{1}{45}\)) + \(\dfrac{1}{41}\)

= 1 - 1 + \(\dfrac{1}{41}\)

= 0 + \(\dfrac{1}{41}\)

= \(\dfrac{1}{41}\)

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

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

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

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

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

\(1,\dfrac{2}{-11}.6\dfrac{2}{7}+\dfrac{3}{7}:3+5\\ =\dfrac{2}{-11}.\dfrac{44}{7}+\dfrac{3}{7}\times\dfrac{1}{3}+5\\ =-\dfrac{88}{77}+\dfrac{3}{21}+5\\ =-\dfrac{8}{7}+\dfrac{1}{7}+5\\ =-1+5\\ =4\)

\(d,1,75:5+2,5.\left(4^2-4.4,1\right)\\ =0,35+2,5.\left(16-16,4\right)\\ =0,35+2,5.0,4\\ =0,35+1\\ =1,35\)

`2,`

`a, 1/2 -x=-1/6`

`=> x= 1/2 -(-1/6)`

`=> x= 1/2 +1/6`

`=> x= 3/6+1/6`

`=>x= 4/6`

`=>x=2/3`

Vậy `x=2/3`

`b, x+3,5 =7,8 -12,6 :2`

`=> x+3,5 = 7,8 - 6,3`

`=> x+3,5=1,5`

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

`=>x=-2`

Vậy `x=-2`

\(c,\dfrac{7}{35}=\dfrac{35-x}{-105}\\ \Rightarrow7.\left(-105\right)=35.\left(35-x\right)\\ \Rightarrow-735=1225-35x\\ \Rightarrow-735-1225=-35x\\ \Rightarrow-1960=-35x\\ \Rightarrow x=-1960:\left(-35\right)\\ \Rightarrow x=56\)

Vậy `x=56`

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

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

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

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

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

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

= - \(\dfrac{1}{4}\)

+ d; \(\dfrac{1}{100.99}\) - \(\dfrac{1}{99.98}\) - \(\dfrac{1}{98.97}\) - ... - \(\dfrac{1}{3.2}\) - \(\dfrac{1}{2.1}\)

  = \(\dfrac{1}{100.99}\) - (\(\dfrac{1}{99.98}\) + \(\dfrac{1}{98.97}\) + ... + \(\dfrac{1}{3.2}\) + \(\dfrac{1}{2.1}\))

 = \(\dfrac{1}{100.99}\) - (\(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ... + \(\dfrac{1}{98.97}\) + \(\dfrac{1}{98.99}\))

=  \(\dfrac{1}{100.99}\) - (\(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ...+ \(\dfrac{1}{98}\) - \(\dfrac{1}{97}\) + \(\dfrac{1}{98}\) - \(\dfrac{1}{99}\))

=  \(\dfrac{1}{100.99}\) - (\(\dfrac{1}{1}\) - \(\dfrac{1}{99}\))

=  \(\dfrac{1}{100.99}\) - \(\dfrac{98}{99}\)

= - \(\dfrac{9799}{9900}\)