dễ mà :  a, 5 - ( 1997 - 2005 ) +1997

= ( 2005 + 5 ) - ( 1997 - 1997 )

= 2010 - 0

= 2010

a, 5  - (1997 -2005 ) + 1997 

= 5 - 1997 + 2005 +1997

=  (5 + 2005 ) - ( 1997 - 1997)

= 2010 - 0 

= 2010

b, 4567 + (1234 - 4567 )-4

=  4567  + 1234 - 4567  -4

= (4567 - 4567 ) + (1234 - 4)

= 0 + 1230 

= 1230 

c, 173 - ( 36 + 173 ) + [(175 - 27 ) - 175 - ( -36 + 50 )]

= 173 - 36 - 173 + 175 - 27 - 175 - 36 - 50 

= ( 173 - 173 ) + (175 - 175 ) - ( 36 + 36 ) - (27 + 50)

= 72 - 77 

= -5

d, 3 . (25 : 5 - 14 : 2 ) - 5 . ( 6 : 2)

= 3  (-2)  - 5 .3

= 3 ( -2 - 5 )

= 3 . (-7 )

= - 21

a) 5-(1997-2005)+1997

= 5-(-1997)+2005+1997

= (5+2005)+(-1997+1997)

= 2010+0

= 0

b) 4567+(1234-4567)-4

 làm tương tự câu a

bài 1 : Tính

2048 : 2³ = 2048 : 8 = 256

bài 1.1 : Tính

625 : 5² . 40 = 625 : 25 . 40 = 25 . 40 = 1000

Bài 2: 

a: \(=3^{11}\)

giúp mk vs nha , mk đang rất cần

a) (38 - 60) + (20 - 38)

= 38 - 60 + 20 - 38

= (38 - 38) + (-60 + 20)

= 0 - 40

= -40

b) 75 - (20 + 75)

= 75 - 20 - 75

= (75 - 75) - 20

= 0 - 20

= -20

c) 32 + (60 - 32)

= 32 + 60 - 32

= (32 - 32) + 60

= 0 + 60

= 60

d) (81 - 36) - (81 - 36)

= 81 - 36 - 81 + 36

= (81 - 81) + (-36 + 36)

= 0 + 0

= 0

e) (2 + 4 + 6 + 8) - (1 + 3 + 5 + 7)

= 2 + 4 + 6 + 8 - 1 - 3 - 5 - 7

= (2 - 1) + (4 - 3) + (6 - 5) + (8 - 7)

= 1 + 1 + 1 + 1

= 4

f) (1 + 3 + 5 + 7 + ... + 99) - (2 + 4 + 6 + 8 + ... + 100)

= 1 + 3 + 5 + 7 + ... + 99 - 2 - 4 - 6 - 8 - ... - 100

= (1 - 2) + (3 - 4) + (5 - 6) + (7 - 8) + ... + (99 - 100)

= -1 - 1 - 1 - 1 - ... - 1 (50 chữ số 1)

= -50

giúp mình với huhuhuhuhuhu

cảm ơn bạn nhiều !!!!!!!!!✿❤☺

(325 - 47) + (175 - 53)

=325 - 47 + 175 -53

=(325 + 175 ) - (47 + 53)

= 500 - 100

= 400
 

b)(756-217)-(183-44)

=756-217-183+44

=(756+44)-(217+183)

=800-400

=400

a)(325-47)+(175-43)

=325-47+175-53

=(325+175)-(47+53)

=500-100

=400

bài 1 tính nhanh

mik xin sửa đề câu a thành thế này ~

\(a,\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

 \(A\cdot2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\) 

\(A\cdot2-A=\) (  \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\) )  - (  \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\) )

\(A=1-\frac{1}{256}\)

\(A=\frac{255}{256}\)

\(b,\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)

đặt \(B=\) \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\) 

     \(B\cdot3=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

\(B\cdot3-B=\)  ( \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)) - \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\) ) 

\(B\cdot2=\) \(1-\frac{1}{729}\)

\(B\cdot2=\frac{728}{729}\)

\(B=\frac{728}{729}:2\)

\(B=\frac{364}{729}\) 

\(c,\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)

ĐẶT \(C=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)

    \(C=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)

\(C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(C=\frac{1}{1}-\frac{1}{6}\)

\(C=\frac{5}{6}\)

Cảm ơn bạn nhé

Bài 2:

a. $=62-81-12+59-9=(62-12)+(59-9)-81$

$=50+50-81=100-81=19$

b. $=39+13-26-62-39=(39-39)+13-(26+62)$

$=0+13-88=-(88-13)=-75$

c. $=(32-42)+(36-34)+(40-38)=10+2+2=14$
d. $=92-55+8-45=(92+8)-(55+45)=100-100=0$

Bài 1:

a. $=(387-87)-224=300-224=76$

b. $=-(75+35)+379=-110+379=379-110=269$

c. $=(11+15)-(13+17)=25-30=-5$

d. $=(31-21)-(27-24)=10-3=7$