`C(y)=(y+2)^2+(y-5)^2=y^2+4y+4+y^2-10y+25=2y^2-6y+29=2(x^2-3x+9/4)+49/2=2(x-3/2)^2+49/2>=49/2`
Dấu "=" xảy ra `<=>x=3/2`
Vậy `C(y)_(min)=49/2<=>x=3/2`
\(C\left(y\right)\)\(=y^2+4y+4+y^2-10y+25=2y^2-6y+29\)
\(=2\left(y-\dfrac{3}{2}\right)^2+\dfrac{49}{2}\ge\dfrac{49}{2}\)
Vậy min \(C\left(y\right)=\dfrac{49}{2}\), đạt được khi và chỉ khi \(y=\dfrac{3}{2}\)