Question

biết \(x^2+3y^2-2x+12y+13\) . khi đó x=... y= .....

Answer 1

Ta có:x²+3y²-2x+12y+13=0<=>x²+3y²-2x+12y+1+12=0

<=>(x²-2x+1)+(3y²+12y+12)=0<=>(x-1)²+3(y+2)²=0

Vì (x-1)²\(\ge0\);3(y+1)²\(\ge0\) nên:(x-1)²+3(y+2)²\(\ge0\)

Dấu "=" xảy ra khi:\(\begin{cases} (x-1)^2=0\\ 3(y+2)^2=0 \end{cases}\)<=>\(\begin{cases} x-1=0\\ y+2=0 \end{cases}\)<=>\(\begin{cases} x=1\\ y=-2 \end{cases}\)

Vậy x=1;y=-2

Answer 2

xin loi mk ghi thieu

x²+3y²-2x+12y+13=0