Giải SBT Toán 7 Bài 4. Quy tắc dấu ngoặc và quy tắc chuyển vế có đáp án

a) (−38)+(79−58)=(−38)+79−58

=(−38)−58+79=−1+79

=−99+79=−29

b) [(−78):2116]−53.(13−710)

=[(−78).1621]−53.(1030−2130)

=(−7) . 168 . 21−53.(−1130)

=−7 . 8 . 28 . 7 . 3+5 . 113 . 5 . 6

=−23+1118=−118

c) [(−23)+34]2.125−15

=[(−812)+912]2.125−15

=(112)2.125−15

=1122.125−15

=160−15=−1160

a) (−0,5)−(−1+23):1,5+(−14)

=(−12)−(−33+23):32+(−14)

=(−12)−(−13).23+(−14)

=(−12)+29+(−14)

=(−1836)+836+(−936)=−1936

b) 49−(37+29)=49−37−29

=49−29−37=29−37

=1463−2763=−1363

c) [(−25)+13]−(35−14)=(−25)+13−35+14

=(−25)−35+13+14=(−1)+13+14

=(−1212)+412+312=−512

d) (112−34)−(0,25+12)=(32−34)−(14+12)

=(32−34)−(14+12)

=(32−12)−(34+14)

= 1 – 1 = 0.

a) A=(8−23+12)−(5−73−32)−(53+52+4)

=(486−46+36)−(306−146−96)−(106+156+246)

=476−76−496=−96=−32

b) A=(8−23+12)−(5−73−32)−(53+52+4)

=8−23+12−5+73+32−53−52−4

=(8−5−4)+(73−23−53)+(12+32−52)

=(−1)+0−12=−32

d) (125−0,4)2:9125−[(113−25).37]

=(125−25)2.1259−[(43−25).37]

=(−925)2.1259−[1415.37]

=92252.1259−25=(32)2(52)2.5332−25

=3454.5332−25=325−25

=95−25=75

a) x+37=25

x=25−37

x=1435−1535

x=−135

b) 32−x=45

x=32−45

x=1510−810

x=710

c) 59−13x=23

13x=59−23

13x=−19

x=(−19):13

x=−13

d) 35x−115=−314:57

355x−65=−310

35x=−310+65

35x=910

x=32

a) 34+19:x=0,5

19:x=12−34

19:x=−14

x=19:(−14)

x=19 . (−4)

x=−49

 

 

 

 

b) 34−(x−23)=113

x−23=34−43

x−23=−712

x=−712+23

x=112

c) (57−x).1115=−2245

57−x=(−2245):1115

57−x=(−2245).1511

57−x=−23

x=57−(−23)

x=2921

d) (2,5x−47):821=−1,5

(52x−47):821=−32

52x−47=(−32).821

52x−47=−47

52x=(−47)+47

52x=0

x=0

a) 1223.713+1123.713

=713 . (1223+1123)

=713 . 1=713

b) 49.2311−111.49+49

=49.(2311−111+1)

=49.(2+1)=49.3=43

c) [(−57)+35]:20202021+(25−27):20202021

=[(−57)+35].20212020+(25−27).20212020

=[(−57)+35+25−27].20212020

=(1−1).20212020=0.20212020=0

d) 38:(722−211)+38:(25−110)

=38:(722−422)+38:(410−110)

=38:322+38:310

=38.(223+103)=38.323=4