Giải thích các bước giải:
`4x^2+25y^2-10x+60y+100`
`=[(2x)^2-2.2x. 5/2+25/4]+[(5y)^2+2.5y.6+36]+231/4`
`=(2x-5/2)^2+(5y+6)^2+231/4`
Có $\left\{\begin{matrix}(2x-\dfrac{5}{2})^2\geq0\\(5y+6)^2\geq0\end{matrix}\right.$
`=>(2x-5/2)^2+(5y+6)^2+231/4>=231/4`
Dấu "=" xảy ra `<=>`$\left\{\begin{matrix}2x-\dfrac{5}{2}=0\\5y+6=0\end{matrix}\right.$
`=>`$\left\{\begin{matrix}2x=\dfrac{5}{2}\\5y=-6\end{matrix}\right.$`=>`$\left\{\begin{matrix}x=\dfrac{5}{4}\\y=-\dfrac{6}{5}\end{matrix}\right.$
Vậy biểu thức `Mi n=231/4<=>`$\left\{\begin{matrix}x=\dfrac{5}{4}\\y=-\dfrac{6}{5}\end{matrix}\right.$
