Răspuns :

Răspuns:

x³+y³-2xy=25

x²-xy+y²=7

a doua ecuatie o inmultim cu (x+y)

x²-xy+y²=7 ⇔(x+y)(x²-xy+y²)=7(x+y) ⇔x³-x²y+xy²+x²y-xy²+y³=7(x+y) ⇔x³+y³=7(x+y)

prima ecuatie devine: x³+y³-2xy=25 ⇔ 7(x+y)-2xy=25

a doua ecuatie devine: x²-xy+y² = 7 ⇔(x²+y²+2xy)-2xy -xy=7⇔ (x+y)²-3xy =7

notam a = x+y si b=xy si inlocum in cele doua ecuatii

⇒ 7a-2b=25 si a²-3b=7

7a-2b=25 ⇒ b= (25-7a)/2

inlocuim b in ecuatia a²-3b=7 ⇔a²-3·(25-7a)/2=7⇔2a²-3·25+3·7a=7·2 ⇔

⇔2a²-75+21a=14 ⇔2a²+21a+61=0 - ecuatie de grad 2

calculam Δ pentru ecuatia 2a²+21a+61=0

Δ=21²-4·2·61=441-488=-47<0 ⇒ ecuatia nu are solutii reale a=x+y

⇒ x+y ∉ R ⇒ x,y∉R

⇒ sistemul de ecuatii in x si y nu are solutii reale