Ta có:

\(\frac{2}{3}+\frac{4}{21}+\frac{6}{91}+\frac{8}{273}+\frac{3}{504}\)

\(=\frac{2}{1.3}+\frac{4}{3.7}+\frac{6}{7.13}+\frac{8}{13.21}+\frac{3}{21.24}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-....+\frac{1}{21}-\frac{1}{24}\)

\(=1-\frac{1}{24}=\frac{23}{24}\)

Vậy giá trị biểu thức là \(\frac{23}{24}\)

A = 4 . 2 . 25 . 5 . 175 

A = 2² . 2 . 5² . 5 . 5² . 7 

A = 2²⁺¹ . 5²⁺¹⁺² . 7  

A = 2³ . 5⁵ .7

A = 175000

\(B=4^2-10^4:\left(50\cdot273-50\cdot73\right)\)

\(B=4^2-10^4:\left[50\cdot\left(273-73\right)\right]\)

\(B=4^2-10^4:\left(50\cdot200\right)\)

\(B=4^2-10^4:10^4=4^2-1=15\)

\(C=3\times53\times6+2\times9\times87-18\times40\)

\(C=18\times53+18\times87-18\times40\)

\(C=18\times\left(53+87-40\right)\)

\(C=18\times100=1800\)

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

(1²+2²...+1000²).(91-273:3)

=(1²+2²...+1000²).(91-91)

=(1²+2²...+1000²).0

=0

tl nhanh hộ mik với

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

\(\left(1^2+2^2+3^2+...+2021^2\right)\cdot\left(91-273:3\right)\)
\(=\left(1^2+2^2+3^2+...+2021^2\right)\cdot\left(91-91\right)\)
\(=\left(1^2+2^2+3^2+...+2021^2\right)\cdot0\)
\(=0\)

(12+22+32+...+20212)⋅(91−273:3)
=(12+22+32+...+20212)⋅(91−91)=(12+22+32+...+20212)⋅(91−91)
=(12+22+32+...+20212)⋅0=(12+22+32+...+20212)⋅0
$=0$

vrrtdtrewrfds5tyhytffedthygfdjhgsawesrdg

a) \(\left(1^2+2^2+3^2+....+2012^2\right).\left(91-273:3\right)\)

\(=\left(1^2+2^2+3^2+...+2012^2\right).\left(91-91\right)\)

\(=0\)

b) \(\left(-284\right).172+\left(-284\right).\left(-72\right)=\left(-284\right).\left(172+-72\right)\)

                                                                             \(=\left(-284\right).100\)

                                                                               \(=-28400\)

c) \(\frac{1}{5}+\frac{-1}{6}+\frac{1}{7}+\frac{-1}{8}+\frac{1}{9}+\frac{1}{8}+\frac{-1}{7}+\frac{1}{6}+\frac{-1}{5}\)

\(=\left(\frac{1}{5}+\frac{-1}{5}\right)+\left(\frac{1}{6}+\frac{-1}{6}\right)+\left(\frac{1}{7}+\frac{-1}{7}\right)+\left(\frac{1}{8}+\frac{-1}{8}\right)+\frac{1}{9}\)

\(=0+0+0+0+\frac{1}{19}\)

= 0

c) Mình nhầm: \(\frac{1}{9}\)

a) \(\left(1^2+2^2+3^2+...+2012^2\right).\left(91-273:3\right)\)

 =   \(\left(1^2+2^2+3^2+...+2012^2\right).0\)

=                               \(0\)

b) (-284) . 172 + (-284) . (-72)

= (-284) . [172 + (-72)]

= (-284) . 100

= -28400

c) \(\frac{1}{5}+\frac{-1}{6}+\frac{1}{7}+\frac{-1}{8}+\frac{1}{9}+\frac{1}{8}+\frac{-1}{7}+\frac{1}{6}+\frac{-1}{5}\)

= \(\frac{1}{9}\)