bỏ ngoặc rồi tính a)(25+38+12)+(2022-63-12) b, (97+43-15) -(87+33-25) C, -(-24+63+71) -(14-73-81)

2 bình luận về “bỏ ngoặc rồi tính a)(25+38+12)+(2022-63-12) b, (97+43-15) -(87+33-25) C, -(-24+63+71) -(14-73-81)”

1. $a)(25+38+12)+(2022-63-12)^{}$

   = $25 + 38 + 12 + 2022 – 63 – 12^{}$

   = $12 – 12  + 25 + 38 – 63 + 2022^{}$

   =  $25 + 38 – 63 + 2022^{}$

   =  $2022^{}$

   $b)(97+43-15) -(87+33-25)^{}$

   $=97+43-15 -87-33+25^{}$

   $=43 – 33 + 97 – 87+ 25 – 15^{}$

   $=10 + 10 + 10^{}$

   $=30^{}$

   c) 

   $-(-24+63+71) -(14-73-81)^{}$

   $=24 – 63 – 71 – 14 + 73 + 81^{}$

   $=(24 – 14) + (63 + 73) +(- 71 + 81)^{}$

   $=10 + 10 + 10^{}$

   $=30 ^{}$

    

   Trả lời
2. a)

   (25 + 38 + 12) + (2022 – 63 – 12)

   = 25 + 38 + 12 + 2022 – 63 – 12

   = (25 + 38 – 63) + (12 – 12) + 2022

   = (63 – 63) + 0 + 2022

   = 0 + 0 + 2022

   = 2022

   b)

   (97+43-15)-(87+33-25)

   = 97+43-15-87-33+25

   = (97-87)+(43-33)+(-15+25)

   = 10+10+10

   = 30

   c)

   -(-24+63+71) – (14 – 73 – 81)

   = 24 – 63 – 71 – 14 + 73 + 81

   = (24 – 14) – (63 – 73) – (71 – 81)

   = 10 + 10 + 10

   = 30

   #BadMo od

   Trả lời