Răspuns :
Efectiv se inmultesc asa cum am invatat (?....) la inmultirea numerelor reprezxentate prin litere tinad cntde regula semnelorla inmultire side reducerea (restrangerea) tetrmenilor asemenea (adica avand acelasi grad in a si in b)
(a+b)²=(a+b)(a+b)=a(a+b) +b(a+b)= a²+ab+ba+b²=a²+ab+ab+b²=a²+2ab+b²
(a-b)²=(a-b)(a-b)=a(a-b)-b(a-b)=a²-ab-ba+(-b)*(-b)=a²-2ab+b²
(a-b)(a+b)=a(a+b)-b(a+b)=a²+ab-ba+(-b)*b=a²-b²
!!!!!!!!!!!!!!!!!!!!!!!!!!!!! toatecelede pan aici trebuiesc invatate si tinute minte PE DE ROST..Se dau la Ev Nationale
urmatoarele doua sunt doar aplicatii....deocamdata...:::)))
(a+b)³= (a+b)(a+b)²=(a+b) (a²+2ab+b²)=a(a²+2ab+b²)+b(a²+2ab+b²)=
=a³+2a²b+ab²+a²b+2ab²+b³=
=a³+3a²b+3ab²+b³=
a³+3ab(a+b)+b³=
a³+b³+3ab(a+b)
(a-b)³=(a-b) (a²-2ab+b²)= a(a²-2ab+b²)-b(a²-2ab+b²)=
=a³-2a²b+ab²-a²b+2ab²-b³=
=a³-3a²b+3ab²-b³
=a³-b³-3ab(a-b)
C.C.T.D.
(a+b)²=(a+b)(a+b)=a(a+b) +b(a+b)= a²+ab+ba+b²=a²+ab+ab+b²=a²+2ab+b²
(a-b)²=(a-b)(a-b)=a(a-b)-b(a-b)=a²-ab-ba+(-b)*(-b)=a²-2ab+b²
(a-b)(a+b)=a(a+b)-b(a+b)=a²+ab-ba+(-b)*b=a²-b²
!!!!!!!!!!!!!!!!!!!!!!!!!!!!! toatecelede pan aici trebuiesc invatate si tinute minte PE DE ROST..Se dau la Ev Nationale
urmatoarele doua sunt doar aplicatii....deocamdata...:::)))
(a+b)³= (a+b)(a+b)²=(a+b) (a²+2ab+b²)=a(a²+2ab+b²)+b(a²+2ab+b²)=
=a³+2a²b+ab²+a²b+2ab²+b³=
=a³+3a²b+3ab²+b³=
a³+3ab(a+b)+b³=
a³+b³+3ab(a+b)
(a-b)³=(a-b) (a²-2ab+b²)= a(a²-2ab+b²)-b(a²-2ab+b²)=
=a³-2a²b+ab²-a²b+2ab²-b³=
=a³-3a²b+3ab²-b³
=a³-b³-3ab(a-b)
C.C.T.D.