a, \(m^2-6m+x^2-x+3\)
\(=m^2-3m-3m+9+x^2-\dfrac{1}{2}x-\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{25}{4}\)
\(=\left(m-3\right)^2+\left(x-\dfrac{1}{2}\right)^2-\dfrac{25}{4}\)
Với mọi giá trị của \(m;x\in R\) ta có:
\(\left(m-3\right)^2+\left(x-\dfrac{1}{2}\right)^2-\dfrac{25}{4}\ge-\dfrac{25}{4}\)
Để \(\left(m-3\right)^2+\left(x-\dfrac{1}{2}\right)^2-\dfrac{25}{4}=-\dfrac{25}{4}\) thì
\(\left\{{}\begin{matrix}\left(m-3\right)^2=0\\\left(x-\dfrac{1}{2}\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m=3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy=..
b, \(3x^2-6x+12\)
\(=3x^2-3x-3x+3+9\)
\(=3x\left(x-1\right)-3\left(x-1\right)+9\)
\(=3\left(x-1\right)^2+9\)
Với mọi giá trị của \(x\in R\) ta có:
\(3\left(x-1\right)^2+9\ge9\)
Để \(3\left(x-1\right)^2+9=9\) thì
\(\left(x-1\right)^2=0\Rightarrow x=1\)
Vậy=..
Chúc bạn học tốt!!!