khi bỏ dấu ngoặc biểu thức 120+(-8)-(42-57) ta được:
A.120+8-42-57
B.120-8-42-57
C.120-8-42+57
D.120+8-42+57
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d) Ta có: \(-8+\left(-14\right)+34+\left(-12\right)\)
\(=-8-14+34-12\)
\(=-20+20\)
\(=-0\)
e) Ta có: \(\left(-7\right)+\left(-20\right)+57+\left(-30\right)\)
\(=\left(-7+57\right)+\left(-20-30\right)\)
\(=50-50=0\)
f) Ta có: \(300-\left(-180\right)-120-42\)
\(=300+180-120-42\)
\(=258+60=318\)
d, Ta có : \(\left(-8\right)+\left(-14\right)+34+\left(-12\right)\)
\(=-\left(8+12\right)+34-14=-20+20=0\)
e, Ta có : \(\left(-7\right)+\left(-20\right)+57+\left(-30\right)\)
\(=-\left(30+20\right)+57-7=-50+50=0\)
f, Ta có : \(300-\left(-180\right)-120-42=318\)
a)58(-45+-24)+-69.42
=58.(-69)+(-69).42
=-69.(58+42)
=-69.100
=-6900
a;
A= 58.(-45) + (-58).24 + (-69).42
A = -58.(45 + 24) - 69.42
A =-58.69 - 69.42
A = -69.(58 + 42)
A = -69.100
A = -6900
b; 139 - (-65 + 239) - (85 + 120)
B = 139 + 65 - 239 - 85 - 120
B = -(239 - 139) - (85 - 65) - 120
B = - 100 - 20 - 120
B = - 120 - 120
B = - 240
Bài 2:
a: \(\dfrac{1}{4};\dfrac{2}{5}\)
\(\dfrac{1}{4}=\dfrac{1\cdot5}{4\cdot5}=\dfrac{5}{20}\)
\(\dfrac{2}{5}=\dfrac{2\cdot4}{5\cdot4}=\dfrac{8}{20}\)
\(\dfrac{2}{3};\dfrac{7}{8}\)
\(\dfrac{2}{3}=\dfrac{2\cdot8}{3\cdot8}=\dfrac{16}{24}\)
\(\dfrac{7}{8}=\dfrac{7\cdot3}{8\cdot3}=\dfrac{21}{24}\)
\(\dfrac{3}{4};\dfrac{5}{6}\)
\(\dfrac{3}{4}=\dfrac{3\cdot3}{4\cdot3}=\dfrac{9}{12}\)
\(\dfrac{5}{6}=\dfrac{5\cdot2}{6\cdot2}=\dfrac{10}{12}\)
b: \(\dfrac{1}{3};\dfrac{7}{9}\)
\(\dfrac{1}{3}=\dfrac{1\cdot3}{3\cdot3}=\dfrac{3}{9}\)
\(\dfrac{7}{9}=\dfrac{7\cdot1}{9\cdot1}=\dfrac{7}{9}\)
\(\dfrac{3}{4};\dfrac{9}{24}\)
\(\dfrac{3}{4}=\dfrac{3\cdot6}{4\cdot6}=\dfrac{18}{24}\)
\(\dfrac{9}{24}=\dfrac{9\cdot1}{24\cdot1}=\dfrac{9}{24}\)
\(\dfrac{7}{10};\dfrac{19}{30}\)
\(\dfrac{7}{10}=\dfrac{7\cdot3}{10\cdot3}=\dfrac{21}{30}\)
\(\dfrac{19}{30}=\dfrac{19\cdot1}{30\cdot1}=\dfrac{19}{30}\)
Bài 1:
\(\dfrac{36}{108}=\dfrac{36:36}{108:36}=\dfrac{1}{3}\)
\(\dfrac{28}{30}=\dfrac{28:2}{30:2}=\dfrac{14}{15}\)
\(\dfrac{42}{98}=\dfrac{42:14}{98:14}=\dfrac{3}{7}\)
\(\dfrac{15}{120}=\dfrac{15:15}{120:15}=\dfrac{1}{8}\)
\(\dfrac{84}{364}=\dfrac{84:28}{364:28}=\dfrac{3}{13}\)
\(\dfrac{120}{100}=\dfrac{120:20}{100:20}=\dfrac{6}{5}\)
\(\dfrac{418}{38}=\dfrac{418:38}{38:38}=\dfrac{11}{1}=11\)
\(\dfrac{96}{1056}=\dfrac{96:96}{1056:96}=\dfrac{1}{11}\)
\(\dfrac{3838}{4040}=\dfrac{3838:101}{4040:101}=\dfrac{38}{40}=\dfrac{38:2}{40:2}=\dfrac{19}{20}\)
\(\dfrac{119119}{123123}=\dfrac{119119:1001}{123123:1001}=\dfrac{119}{123}\)
1) 25 + 91 + 75 + 101
= (25+75) + (91 + 101)
= 100 + 192
= 292
2) 41 x 39 - 41 x 9
= 41 x (39 - 9)
= 41 x 30
= 1230
3) 710 x 99 + 710
= 710 x 99 + 710 x 1
= 710 x (99 + 1)
= 710 x 100
= 71000
4) 17 x 85 + 15 x 17 - 120
= 17 x (85 + 15) - 120
= 17 x 100 -120
= 1700 -120
= 1580
5) (39 x 42 - 37 x 42) : 42
= 42 x (39 - 37) : 42
= 42 x 2 : 42
= 84 : 42
=2
6) 120 - 48 - 52
= 120 - (48 + 52)
= 120 - 100
= 20
7) 23 x 75 + 25 x 23 + 180
= 23 x (75 + 25) +180
= 23 x (75 + 25) +180
= 23 x 100 + 180
= 2300 +180
=2480
8) 2448 : [119 - (23 - 6)]
= 2448 : (119 - 17)
= 2448 : 102
= 24
A) ( 42 x 43 + 43 x 57 + 43 ) - 360 : 4
= ( 42 x 43 + 43 x 57 + 43 x 1 ) - 360 : 4
= [ 43 x ( 42 + 57 + 1 ) ] - 360 : 4
= ( 43 x 100 ) - 360 : 4
= 4300 - 360 : 4
= 4300 - 90
= 4210
B) 322 - 19 x 4 + ( 981 : 9 - 13 )
= 322 - 19 x 4 + ( 109 - 13 )
= 322 - 19 x 4 + 96
= 322 - 76 + 96
= 246 + 96
= 342
C) 450 : 2 x 18 + 456 : 3 - 120
= 225 x 18 + 152 - 120
= 4050 + 152 - 120
= 4202 - 120
= 4082
C
khi bỏ dấu ngoặc biểu thức 120+(-8)-(42-57) ta được:
A.120+8-42-57
B.120-8-42-57
C.120-8-42+57
D.120+8-42+57