Question

dau coroana URGENT!!!!!!!!!!!!!

Answer 1

 

$\displaystyle\\ 12=1+9+2\\\\4x^2+y^2-4x+6y+12=\\\\=4x^2-4x+y^2+6y+1+9+2=\\\\=4x^2-4x+1+y^2+6y+9+2=\\\\=(4x^2-4x+1)+(y^2+6x+9)+2=\\\\=(2x-1)^2+(y+3)^2+2\\\\(2x-1)^2\geq0\\\\(y+3)^2\geq0\\\\2>0\\\\\implies~~\boxed{(2x-1)^2+(y+3)^2+2\geq0}$

Answer 2

Răspuns:

Explicație pas cu pas:

4x2-4x+3+y2+6y+9 >2

4(x2-x+3/4)

x2-x+3/4  are minima (y"=2,resp>0)  in m(xm,ym)

y'=2xm-1=0  => xm=1/2 => ym= (1/2)^2-1/2+3/4=1/4-2/4+3/4=2/4=1/2

4*(1/2) >0

(y+3)^2  =0  pt y=-3