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David2511
a fost răspuns

x+y=17,5 si y+z=14,1 calculati (x+3y+2z)×(z minus x)

Răspuns :

x+3y-3z=x+2y+y-3z=17+4=21
2x+5y-3z=2x+4y+y-3z=2(x+2y)+y-3z=2*17+4=34+4=38
(3x+6y)+(2y-6z)=3(x+2y)+2(y-3)=3*17+2*4=51+8=59
(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)=34*(17+8)=34*25=850

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