a) A = 37. 173 +62. 173 + 173

= 173. (37 +62 + 1)

= 173.100

= 17 300

b) B = 72.99 + 28.99 - 900

= 99. (72+28) - 900

= 99.100 - 900

= 9 900 - 900

= 9 000

c) C = 2³.3-(1¹⁰ + 15): 4²

= 8.3 - (1 + 15): 4²

= 8.3 - 16:4²

= 8.3 - 1

= 23.

d) D = 6²: 4.3 + 2.5² - 201⁰

= 36:4.3 + 2.25 - 1

= 27 + 50 - 1

= 76.

a) 128 - [(1317 - 1257) . 90 + 61] : 43

= 128 - [60 . 90 + 61] : 43

= 128 - [5400 + 61] : 43

= 128 - 5461 : 43

= 128 - 127

= 1

b) 720 : {4320 : [427 - (173 + 25 . 8)]}

= 720 : {4320 : [427 - (173 + 200)]}

= 720 : {4320 : [427 - 373]}

= 720 : {4320 : 54}

= 720 : 80

= 9.

Đáp án là B

5) 327 - 48 - 152 + 173 =(327+173)-(48+152)=500-200=300

6) A = \(\frac{154+919+146+781}{823-217+533-139}=\frac{\left(154+146\right)+\left(919+781\right)}{1000}\)

=\(\frac{300+1700}{1000}=\frac{2000}{1000}=2\)

7) ( 1354 - 356 : 6 ) + 13,2=(1354\(-\frac{178}{3}\))+13,2=\(\frac{3884}{3}+13,2=\frac{19618}{15}\)

8) 784 . 0,5 : 0,28 - 196 : 28 + 2003=392:0,28-7+2003

=1400-7+2003=3396

   173 - 536 + 173 - [ 29 - 29 + 526 ]

= 173 - 536 + 173 - 526

= ( 173 + 173 ) - 536 - 526

= 346 - 536 - 526

= -716

                                                            Bài giải

\(173-536+173-[29-29+526]\)

\(=(173+173 )-536-[( 29-29)+526]\)

\(=346-536-526\)

\(=346-\left(536+526\right)\)

\(=346-1042\)

\(=-716\)

\(A=\frac{-173.24+42.\left(-173\right)-173.34}{-18.519+\left(-519\right).49+33.\left(-519\right)}=\frac{\left(-173\right).24+42.\left(-173\right)+\left(-173\right).34}{\left(-519\right).18+\left(-519\right).49+33.\left(-519\right)}\)

\(=\frac{\left(-173\right).\left(24+42+34\right)}{\left(-519\right).\left(18+49+33\right)}=\frac{\left(-173\right).100}{\left(-519\right).100}=\frac{-173}{-519}=\frac{1}{3}\)

\(B=\frac{10.11.12+20.22.24+30.33.36+40.44.48}{10.12.14+20.24.28+30.36.42+40.48.56}\)

\(=\frac{10.11.12+2.10.2.11.2.12+3.10.3.11.3.12+4.10.4.11.4.12}{10.12.14+2.10.2.12.2.14+3.10.3.12.3.14+4.10.4.12.4.14}\)

\(=\frac{10.11.12+8.10.11.12+27.10.11.12+64.10.11.12}{10.12.14+8.10.12.14+27.10.12.14+64.10.12.14}\)

\(=\frac{\left(10.11.12\right).\left(1+8+27+64\right)}{\left(10.12.14\right).\left(1+8+27+64\right)}=\frac{10.11.12}{10.12.14}=\frac{11}{14}\)

b)(1-3-5+7)+(9-11-13+15)+...+(2009-2011-2013+2015)

=0+0+...+0

=0

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