\(-47.69+31.\left(-47\right)\)

\(=-47\left(69+31\right)\)

\(=-47.100\)

\(=-4700\)

\(25.11.\left(-15\right)-50.15.\left(-5\right)\)

\(=25.11.\left(-15\right)-25.15.2.\left(-5\right)\)

\(=25.\left(-15\right).11-25.\left(-15\right).10\)

\(=25.\left(-15\right).\left(11-10\right)\)

\(=25.\left(-15\right)=-375\)

a, -47.69+31.(-47)

=-47.(69+31)

=-47.100

=-4700
 

a)

-47.69 + 31.(-47)

= (-47)(69+31)

=(-47).100=-4700

b)

25.11.(-15)  - 50.15.(-5)

=25.11.(-15) - 25.2.3.5.(-5)

=25.11.(-15) - 25.10.(-15)

=(-15)(25.11-25.10)

=(-15)[25(11-10)]

=(-15).25=-375

c) 2015.|-15|-2015.|15|-|-3|.2015

= 2015.15-2015.15-3.2015

= (2015-2015).2015.(15-15).3

= 0.2015.0.3

=0

=147/20

Ta có :

A = 24x + 7y - 11x + 6y

=> A = ( 24x - 11x ) + ( 7y + 6y )

=> A = 13x + 13y 

=> A = 13( x + y ) mà x + y = 13

=> A = 13²

Ta có :

B = ( 1 + 2 + 3 + ... + 100 ) . ( 250 . 3 - 50 . 15 )

=> B = ( 1 + 2 + 3 + ... + 100 ) . ( 750 - 750 )

=> B = ( 1 + 2 + 3 + ... + 100 ) . 0

=> B = 0

35.5 - 15.(- 5)

= (35.15).[5 - (- 5)]

= 525.10

= 5250

`35.5-15.(-5)`

`=35.5+15.5`

`=5.(35+15)`

`=5.50`

`=250`

35.5 - 15.(-5)

= 5.( 35 + 15)

= 5 .  50

= 250 

\(\dfrac{5}{13}.\dfrac{8}{15}+\dfrac{5}{13}.\dfrac{26}{15}-\dfrac{5}{13}.\dfrac{8}{15}.\\ =\dfrac{5}{13}.\left(\dfrac{8}{15}+\dfrac{26}{15}-\dfrac{8}{15}\right).\\ =\dfrac{5}{13}.\dfrac{26}{15}=\dfrac{2}{3}.\)

\(\dfrac{5}{13}\times\dfrac{8}{15}+\dfrac{5}{13}\times\dfrac{26}{15}-\dfrac{5}{13}\times\dfrac{8}{15}\)

\(=\dfrac{5}{13}\times\left(\dfrac{8}{15}+\dfrac{26}{15}-\dfrac{8}{15}\right)\)

\(=\dfrac{5}{13}\times\dfrac{26}{15}\)

\(\dfrac{2}{3}\)

Ta có: \(B=\dfrac{5}{13}\cdot\dfrac{8}{15}+\dfrac{5}{13}\cdot\dfrac{26}{15}-\dfrac{5}{13}\cdot\dfrac{8}{15}\)

\(=\dfrac{5}{13}\cdot\dfrac{26}{15}\)

\(=\dfrac{130}{195}=\dfrac{2}{3}\)