(10²+11²+12²):(13²+14²)=(100+121+144):(169+196)=365:365=1

đáp sồ bài này bằng 1

a)\(\left(10^2+11^2+12^2\right)\div\left(13^2+14^2\right)\)

\(=\left(100+121+144\right)\div\left(169+196\right)\)

\(=365\div365\)

\(=1\)

b) \(1.2.3...9-1.2.3...8-1.2.3...8^2\)

\(=1.2.3...8\left(9-1-8\right)\)

\(=1.2.3...8.0\)

\(=0\)

d) \(1152-\left(374+1152\right)+\left(-65+374\right)\)

\(=1152-374-1152-65+374\)

\(=\left(1152-1152\right)-65+\left(374-374\right)\)

\(=0-65+0\)

\(=-65\)

e) \(13-12+11+10-9+8-7-6+5-4+3+2-1\)

\(=13-\left(12-11\right)+\left(10-9\right)+\left(8-7\right)-\left(6-5\right)-\left(4-3\right)\)\(+\left(2-1\right)\)

\(=13-1+1+1-1-1+1\)

\(=13+0+0+0\)

\(=13\)

Bài 1: 
a) 10³⁰ và 2¹⁰⁰
   Ta có: 10³⁰ = (10³)¹⁰ = 1000¹⁰
              2¹⁰⁰ = (2¹⁰)¹⁰ = 1024¹⁰
   Vì 1000¹⁰<1024¹⁰ nên 10³⁰<2¹⁰⁰
   Vậy ...
b) 5⁴⁰ và 620¹⁰
   Ta có: 5⁴⁰= (5⁴)¹⁰ =625¹⁰
   Vì 620¹⁰<625¹⁰ nên 620¹⁰<5⁴⁰
   Vậy...

b/1.2.3.....2013 - 1.2.3....2012-1.2.3...2012²

= (1.2.3...2012.2013-1.2.3...2012)-1.2.3...2012.2012

= 1.2.3...2012(2013-1-2012)

=1.2.3...2012.0=0

kkkkkkkkkkkkk giải giùm mình đi

=(100+121+144):(169+196)

=365:365

=1

1 

chắc chắn 100%

bằng 1 nha bn

( 102 + 112 + 122 )  :  ( 132  + 142 )
= (100 + 121 + 144) : (169 + 196)
= 365 : 365
= 1

Câu b:

B = (3\(^{10}\).11 + 3\(^{10}\).5) : (3\(^9\).2\(^4\))

B = 3\(^{10}\).(11 + 5): (3\(^9\).2\(^4\))

B = 3\(^{10}\).16:(3\(^9\).2\(^4\))

B = (3\(^{10}\) : 3\(^9\)).(16 : 2\(^4\))

B = 3.1

B = 3

Câu c:

C = (2\(^{10}\).13 + 2\(^{10}\).65) : (2\(^8\).104)

C = 2\(^{10}\).(13 + 65):(2\(^8\).104)

C = 2\(^{10}\).78 : (2\(^8\).104)

C = (2\(^{10}\) : 2\(^8\)).(78 : 104)

C = 2\(^2\).\(\frac34\)

C = 3

E= 2⁹.2⁶:2¹³ = 2^9+6-13 = 2² = 4

F=3⁵.27.2⁶ = 3⁵.3³.2⁶ = 3⁸.2⁶ = 419904

G=64.2¹⁰ :2¹⁴ = 2⁶.2¹⁰ :2¹⁴ = 2^6+10-14 = 2² = 4

F=32.128:2¹⁰ = 2⁵.2⁷:2¹⁰ = 2^5+7-10 = 2² = 4

E=4

F=41990

G=4

F=4