a) a/(a-b)(a-c) + b/(b-a)(b-c) + c/(c-a)(c-b)=0
a/(a-b)(a-c) +b/ -(a-b)(b-c) + c/(a-c)(b-c) =0
a/(a-b)(a-c) -b/(a-b)(b-c) + c/(a-c)(b-c) =0
Aduc la acelasi numitor (a-b)(a-c)(b-c) si il elimin
a(b-c) - b(a-c) + c(a-b) =0
ab-ac -ab + bc +ac -bc =0 Adevarat
b) a^2/(a-b)(a-c) + b^2/(b-a)(b-c) + c^2/(c-a)(c-b)=1
sunt exact aceiasi pasi
a^2(b-c) -b^2(a-c) + c^2(a-b)= (a-b)(a-c)(b-c)
a^2*b -a^2*c - b^2*a+ b^2*c + c^2*a- c^2*b= (a^2-ac-ab+bc)(b-c)
a^2*b -a^2*c - b^2*a+ b^2*c + c^2*a- c^2*b=a^2b-abc -ab^2 + b^2c -a^2c +ac^2 +abc-bc^2
a^2*b -a^2*c - b^2*a+ b^2*c + c^2*a- c^2*b=a^2*b -a^2*c - b^2*a+ b^2*c + c^2*a- c^2*b Adevarat
c) 1/a(a-b)(a-c) + 1/ b(b-a)(b-c) + 1/c(c-a)(c-b)=1/abc
noul numitor aici este abc(a-b)(a-c)(b-c)
bc(b-c)- ac(a-c) + ab(a-b) =(a-b)(a-c)(b-c)
b^2c - bc^2 -a^2c +ac^2 +a^2b- ab^2=a^2b-abc -ab^2 + b^2c -a^2c +ac^2 +abc-bc^2
b^2c - bc^2 -a^2c +ac^2 +a^2b- ab^2=a^2b -ab^2 + b^2c -a^2c +ac^2 -bc^2
Adevarat
Am mai sarit niste pasi la b si c pt ca se repeta..sunt aceiasi ca la a
sper sa intelegi
