a) \(\left(29,42+20,58\right)-\left(34,23+25,77\right)\)

\(=50-60=-10\)

b) \(-\left(212,49+87,51\right)+99,9\)

\(=-300+99,9=-200,1\)

A)29.42+20.58−34.23+(−25.77)29.42+20.58−34.23+(−25.77) 

=29.42+20.58−34.23−25.77

=29.42+20.58−34.23−25.77

=(29.42+20.58)−(34.23+25.77)

=(29.42+20.58)−(34.23+25.77)

=50−60=50−60

=−10=−10

B).(−212.49)−(87.51−99.9)(−212.49)−(87.51−99.9) 

=−212.49−87.51+99.9

=−212.49−87.51+99.9

=−(212.49+87.51)+99.9

=−(212.49+87.51)+99.9

=−300+99.9=−300+99.9

=−(300−99.9)=−(300−99.9)

=−200.1

a) 2763 + 152 = 2915

\(2915\)

1x\(10^{17}\)

100000000000000000

cho mik cách giải

\(a.\) \(50\cdot374\cdot2=\left(50\cdot2\right)\cdot347=100\cdot347=34700\)

\(b.\)\(36\cdot97+97\cdot64=97\cdot\left(36+64\right)=97\cdot100=9700\)

\(c.157\cdot289-289\cdot57=289\cdot\left(157-57\right)=289\cdot100=28900\)

\(a)\left(-17\right)+\left(-23\right)+44=-40+44=4\)

\(b)\left(-39\right)+\left(-16\right)+39=\left[\left(-39\right)+39\right]+\left(-16\right)=0+\left(-16\right)=-16\)

a) (- 17) + (- 23) + 44;

= (- 40) + 44

= 4

b) (- 39) + (- 16) + 39.

= (- 39) + 39 + (- 16)

= 0 + (- 16)

= - 16

a) (- 17) + (- 23) + 44

=  [ (- 17) + (- 23) ] + 44 

=  (- 40) + 44

= 4

b) (- 39) + (- 16) + 39

= [ (- 39) + 39 ] + (- 16)

= 0 + (- 16)

= -16

41.29 + 41.70 + 41

= 41.(29 + 70 + 1)

= 41. { (29+1) + 70}

= 41.(30 + 70)

= 41.100

= 4100

= 37.(72 - 5 + 33)

=37.{ (72 + 33) - 5}

= 37. (105 - 5)

= 37. 100

= 3700 

a) 21 - 22 + 23 - 24

= (21 - 22) + (23 - 24)

= (-1) + (-1)  = -2

b) 125 - (115 - 99)

= 125 - 115 + 99

= (125 - 115) + 99

 = 10 + 99 = 109.

Lời giải:

$A=(1-4)+(7-10)+(13-16)+....+(97-100)+103$

$=\underbrace{(-3)+(-3)+(-3)+...+(-3)}_{17}+103$

$=(-3).17+103=52$

\(=\left(-\dfrac{3}{11}-\dfrac{8}{11}\right)+\left(\dfrac{11}{8}-\dfrac{3}{8}\right)=0\)