\(42+\frac{93}{6}+\frac{87}{12}+\frac{79}{20}+\frac{69}{30}+\frac{57}{42}\)

\(=37+\left(1+\frac{93}{6}\right)+\left(1+\frac{87}{12}\right)+\left(\frac{79}{20}+1\right)+\left(1+\frac{69}{30}\right)+\left(1+\frac{57}{42}\right)\)

\(=37+\frac{99}{6}+\frac{99}{12}+\frac{99}{20}+\frac{99}{30}+\frac{99}{42}\)

\(=37+99\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

\(=37+99\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

\(=37+99\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(=37+99\left(\frac{1}{2}-\frac{1}{7}\right)\)

\(=37+99.\frac{5}{14}\)

\(=37+35\frac{5}{14}\)

\(=72\frac{5}{14}\)

Khiếp

2 . 53 . 27 + 6 . 9 . 87 - 3 . 18 . 40 
= 2. 53 . 9 . 3 + 2. 3 .9 .87 - 3 .2 .9 .40
= 54.53+54.87-54.40
=54(53+87-40)
=54. 100=5400

12 . 25 + 29 . 25 + 59 . 25 
= 25 . ( 12+29+59)
=25 . 90
=2250

43 . 37 + 93 . 43 + 57 . 61 + 69 . 57 
= 43.(37+93)+57(61+69)
=43.130+57.130
=130.(43+57)
=130.100
=13 000

12 . 53 + 53 . 172 - 53 
=53(12+172-1)
=53.184
=9752

35 . 54 + 35 . 58 + 65 . 75 + 65 . 45 
=35(54+58)+65(75+45)
=35.112+65.120
=3920+7800
=11720

39 . 8 + 70 . 3 + 31 . 8 
=8(39+31)+70.3
=8.70+70.3
=70.(8+3)
=70.11
=770

2 . 53 . 27 + 6 . 9 . 87 - 3 . 18 . 40

= ( 2 . 27 ) . 53 + ( 6 . 9 ) . 87 - ( 3 . 18 ) . 40

= 54 . 53 + 54 . 87 - 54 . 40

= 54(53+87-40)

= 5400

12 . 25 + 29 . 25 + 59 . 25

= 25( 12 + 29 + 59 )

= 2500

43 . 37 + 93 . 43 + 57 . 61 + 69 . 57

= 43( 37 + 93 ) + 57(61 + 69 )

= 5590 + 7410

= 13000

12 . 53 + 53 . 172 - 53 

= 53( 12 + 172 ) - 53

= 9752 - 53

= 9699

35 . 54 + 35 . 58 + 65 . 75 + 65 . 45

= 35( 54 + 58 ) + 65( 75 + 45 )

= 3920 + 7800

= 11720

Bài 2:

a: \(387+\left(-224\right)+\left(-87\right)\)

\(=\left(387-87\right)+\left(-224\right)\)

=300-224

=76

b: \(39+\left(13-26\right)-\left(62+39\right)\)

\(=39+13-26-62-39\)

\(=\left(39-39\right)+\left(13-26-62\right)\)

=0-75

=-75

c: \(32-34+36-38+40-42\)

\(=\left(32-34\right)+\left(36-38\right)+\left(40-42\right)\)

\(=\left(-2\right)+\left(-2\right)+\left(-2\right)\)

=-6

d: \(92-\left(55-8\right)+\left(-45\right)\)

\(=92-55+8-45\)

\(=\left(92+8\right)-\left(55+45\right)\)

=100-100

=0

e: -20<=x<=20

mà x nguyên

nên \(x\in\left\{-20;-19;-18;...;18;19;20\right\}\)

=>M={-20;-19;-18;-17;...;18;19;20}

Tổng các phần tử của M là:

\(\left(-20\right)+\left(-19\right)+\left(-18\right)+\left(-17\right)+...+18+19+20\)

\(=\left(-20+20\right)+\left(-19+19\right)+...+\left(-2+2\right)+\left(-1+1\right)+0\)

=0+0+...+0

=0

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\)

\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)

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

\(=1-\dfrac{1}{7}\)

\(=\dfrac{6}{7}\)

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\)

\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)

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

\(=1-\dfrac{1}{7}\)

\(=\dfrac{6}{7}\)

A = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72

   = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9

   = 1- 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9

   = 1 - 1/9 = 8/9

Câu B, C dấu * là nhân hay công vậy?

\(\frac{-1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

= \(-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

=\(-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

=\(-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

=\(-\left(1-\frac{1}{10}\right)=-\left(\frac{9}{10}\right)=-\frac{9}{10}\)

\(\frac{-1}{90}-\frac{-1}{72}-\frac{-1}{56}-\frac{-1}{42}-\frac{-1}{30}-\frac{-1}{20}-\frac{-1}{12}-\frac{-1}{6}-\frac{-1}{2}\)

\(=\frac{-1}{10.9}-\frac{-1}{9.8}-\frac{-1}{8.7}-\frac{-1}{7.6}-\frac{-1}{6.5}-\frac{-1}{5.4}-\frac{-1}{4.3}-\frac{-1}{3.2}-\frac{-1}{2.1}\)

\(=\frac{79}{90}\)

1)

\(45\left(13+78\right)+9\left(87+22\right)5\)

\(=45\left(13+78\right)+45\left(87+22\right)\)

\(=45\left(13+78+87+22\right)\)

\(=45\left[\left(87+13\right)+\left(78+22\right)\right]\)

\(=45\left[100+100\right]\)

\(=45.200\)

\(=9000\)

2)

\(43+37+93.43+57.61+69.57\)

\(=43\left(37+93\right)+57.\left(61+69\right)\)

\(=43.130+57.130\)

\(=130\left(43+57\right)\)

\(=130.100\)

=13000

1/100 - 1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2 = -0,833