a)(x+1)²+(y-3)²=0
(x+1)²≥0; (y-3)²≥0
=> x+1=0; x=-1 si y-3=0; y=3
b)x²+2x+y²+6y+10=0
(x²+2x+1)+ (y²+6y+9)=0
(x+1)²+(y+3)²=0
=> x+1=0; x=-1 si y+3=0; y=-3
c)4x²+y²+4x+10y+26=0
(4x²+4x+1)+ (y²+10y+25)=0
(2x+1)²+(y+5)²=0
2x+1=0; x=-1/2 si y+5=0; y=-5
Răspuns:
x+1=0 x=-1 ; y-3=0 ; y=3 ; b) (x+1)²+(y+3)²=0 ; x+1=0 ; x=-1 = y+3=0 ; y=-3 ; (2x+1)²+(y+5)²=0 ; 2x+1=0 2x=1 ; x=1/2 ; y+5=0 y=-5
Explicație pas cu pas: