a)(13+49)-(13-135+49)

=13+49-13+135-49

=13-13+49-49+135

=0+0+135

=135

b)(18+29)+(158-18-29)

=18+29+158-18-29

=18-18+29-29+158

=0+0+158

=158

c)4 .5²-3 .(24-9)

=4. 5²- (3.24-3.9)

=4. 5²-3.24+3.9

=4.25-72+27

=100-72+27

=28+27

=55

d)[93-(20-7)]:16

=(93-20+7):16

=80:16

=5

chúc cậu hok tốt

a) (13 + 49) - (13 - 135 + 49)

= 13 + 49 - 13 + 135 - 49

= (13 - 13) + (49 - 49) + 135

= 0 + 0 + 135

= 135

b) (18 + 29) + (158 - 18 - 29)

= 18 + 29 + 158 - 18 - 29

= (18 - 18) + (29 - 29) + 158

= 0 + 0 + 158

= 158

c) 4.5² -3.(24 - 9)

= 4.25 - 3.15

= 100 - 45

= 55

d) [93 - (20 - 7)] : 16

= [93 - 13] : 16

= 80 : 16

= 5

a,\((13+49)-(13-135+49)\)

\(=13+49-13-135+49\)

\(=(13-13)+(49-49)+135\)

\(=0+0+135\)

\(=135\)

\(b,(18+29)+(158-18-29)\)

\(=18+29+158-18-29\)

\(=(18-18)+(29-29)+158=0+0+158=158\)

c,Tự làm

d,\([93-(20-7)]:16\)

\(=(93-13):16=\frac{80}{16}=5\)

c2:9+(9-9+9/9)=9+1=10

c4:9+9⁹/9⁹=9+1=10

Cách 1: Tính giá trị từng biểu thức trong ngoặc

Cách 2: Bỏ dấu ngoặc rồi nhóm các số thích hợp

C1: 9 + 99 : 99 = 9 + 1 = 10

C2: 9 + (9 - 9 + 9 : 9) = 9 + (0 + 1) = 9 + 1 = 10

C3: 9 + 99^(9-9) = 9 + 99^0 = 9 + 1 = 10

C4: 9 + 9^9 : 9^9 = 9 + 1 = 10

C1 : 9 : 9 + 9 + 9 - 9

C2 : 9+99:99

C3 : 9+(9-9+9:9)

C4 : 9+9^9:9^9

Cách 1:
\(A=\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
  \(=\left(\frac{36}{6}-\frac{4}{6}+\frac{3}{6}\right)-\left(\frac{30}{6}+\frac{10}{6}-\frac{9}{6}\right)-\left(\frac{18}{6}-\frac{14}{6}+\frac{15}{6}\right)\)
  \(=\frac{35}{6}-\frac{31}{6}-\frac{19}{6}\)
  \(=-\frac{15}{6}\)
  \(=-\frac{5}{2}\)

Cách 2:
\(A=\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
  \(=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
  \(=\left(6-5-3\right)+\left(-\frac{2}{3}-\frac{5}{3}+\frac{7}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)
  \(=-2+0-\frac{1}{2}\)
  \(=-\frac{4}{2}-\frac{1}{2}\)
  \(=-\frac{5}{2}\)

C1
23+12)−(5+53−32)−(3−73+52)=(6.66−2.26+36)−(5.66+5.26−3.36)−(3.66−7.26+5.36)=36−4+36−30+10−96−18−14+156=356−316−196=35−31−196=−156=−52=−212.

23+12)−(5+53−32)−(3−73+52)=6−23+12−5−53+32−3+73−52=(6−5−3)+(−23−53+73)+(12+32−52)=−2+−2−5+73+1+3−52=−2+0−12=−52=−212." role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.94px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">A=(6−23+12)−(5+53−32)−(3−73+52)=6−23+12−5−53+32−3+73−52=(6−5−3)+(−23−53+73)+(12+32−52)=−2+

12 - 12 + 11 + 10 - 9 + 8 - 7 + 5 - 4 + 3 + 2 - 1

= ( 12 - 12 ) + 11 + ( 10 - 9 ) + ( 8 - 7 ) + ( 5 - 4 ) + 3 + ( 2 - 1 )

= 0 + 11 + 1 + 1 + 1 + 3 + 1

= 0 + 11 +(1 + 1 + 1 + 3 + 1)

= 11 + 7

= 18