\(\left(1^2+2^2+3^2+...+1000000^2\right).\left(91-273:3\right)\\ =\left(1^2+2^2+3^2+...+1000000^2\right).0=0\)

(1²+2²+3²+..+1000000²).(91-273:3)
=(1²+2²+3²+..+1000000²).(91-91)

=(1²+2²+3²+..+1000000²).0

=0

[1²+ 2²+3²+...+1000 000²].[91 - 273:3]

[1²+2²+3²+....+1000 000²].0 

=0

là các chữ số chẵn

so cuoi cung la 4

(1²+2²...+1000²).(91-273:3)

=(1²+2²...+1000²).(91-91)

=(1²+2²...+1000²).0

=0

a) 19 + (x - 23) = 50

x - 23 = 50 - 19

x - 23 = 31

=> x  = 31 + 23

     x = 54

b) (273 - x) - 140 = 34

(273 - x) = 34 + 140

(273 - x) = 174

x = 273 - 174

x = 99

c) 56 - (x - 14) = 22

x - 14 = 56 - 22

x - 14 = 34

x = 34 + 14

x = 48

d) 172 - (137 - x) = 93

137 - x = 172 - 93

137 - x = 79

x = 137 - 79

x = 58

e) 13. (x - 14) = 91

x - 14 = 91 : 13

x - 14 = 7

x = 14 + 7

x = 21

g) (13 - x).17 = 51

13 - x = 51 : 17

13 - x = 3

x = 13 - 3

x = 10

chúc bn học giỏi ^^ !

ok mk nha!! ! 5654645645767676576558578779769675634636457567687689978978978978766876765657886567567

1.19+(x-23)=50

<=>x-23=50-19

<=>x-23=31

<=>x=31+23

<=.x=54

vậy x=54

2.(273-x)-140=34

<=>273-x=34+140

<=>273-x=174

<=>x=273-174

<=>x=99(thoả mãn)

vậy x=99

a) \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^{32}-1\right)< 3^{32}-1=B\)

b) \(A=2011.2013=\left(2012-1\right)\left(2012+1\right)=2012^2-1< 2012^2=B\)

a: =1/8-1/8+2/5=2/5

b: =(-1/15-14/15)+(23/27+31/27)=2-1=1

c: \(=\dfrac{3}{7}\left(\dfrac{22}{21}+\dfrac{5}{21}+\dfrac{15}{21}\right)=\dfrac{3}{7}\cdot2=\dfrac{6}{7}\)

d: \(=\dfrac{-8}{9}\cdot\dfrac{3}{2}+\dfrac{1}{9}\cdot\dfrac{-3}{2}=-\dfrac{3}{2}\)