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

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

=(1²+2²+3²+..+1000000²).0

=0

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

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

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

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

a)  >           

b) <           

c) >            

d) >

e) >            

f) <             

g) >            

h) <

1)-73

2)-5

3)-64

4)-56

5)9

6)-20

7)6

8)-2

Tick nha

mk ra kết quả thôi nha

(1²+2²...+1000²).(91-273:3)

=(1²+2²...+1000²).(91-91)

=(1²+2²...+1000²).0

=0

Ta có 1² + 2² + 3² + …10² = 385

Suy ra ( 1² +2² + 3² +…+10² ) .3² = 385.3²

Do đó ta tính được A = 3² + 6² + 9² + …+30²  = 3465

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 \(\frac{23}{24}\)