Question

tot la matematica am niste necunoscute.
daca x+2y=17 si y-3z=4, calculati:
a) x+3y-3z
b) 2x+5y-3z
c. (3z+6y)+(2y-6z)
d. (2x+4y)*(x+4y-6z)

Answer 1

a) x+3y-3z= x+2y+y-3z si de aici aflam 17 + 4=21

Answer 2

a) x+3y-3z= (x+2y)+(y-3z)= 17+4=21

b) 2x+5y-3z=

c) (3x+6y)+(2y-6z)= 3(x+2y)+2(y-3z)= 3*17+3*4= 51+12=63

d) (2x+4y)*(x+4y-6z)=2(x+2y)*(x+2y+2y-6z)= 2(x+2y)*(x+2y)2(y-3z)=2*17*17*2*4=4624

cred ca este bine