a.A= \(x^2+10x+27\)
\(=x^2+2.x.5+25+2\)
\(\left(x+5\right)^2+2\ge2\forall x\)
Dấu " = " xảy ra <=> x + 5 = 0
=> x = -5
Vậy Min A = 2 <=> x = -5
b.B = \(x^2-12x+37\)
\(=x^2-2.x.6+36+1\)
\(=\left(x-6\right)^2+1\ge1\forall x\)
Dấu " = " xảy ra <=> x - 6 = 0
=> x = 6
Vậy Min B = 1 <=> x = 6
c. \(x^2+x+7\)
\(=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{27}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}\forall x\)
Dấu " =" xảy ra <=> \(x+\dfrac{1}{2}=0\)
\(x=\dfrac{-1}{2}\)
Vậy Min C = \(\dfrac{27}{4}\Leftrightarrow x=\dfrac{-1}{2}\)
\(F=x^2+4xy+2y^2-22y+173\)
\(=\left(x^2+2xy+y^2\right)+\left(y^2-22y+121\right)+52\)
\(=\left(x+y\right)^2+\left(y-11\right)^2+52\)
\(\left(x+y\right)^2\ge0;\left(y-11\right)^2\ge0\forall x,y\)
\(\Rightarrow F\ge52\forall x,y\)
Dấu " =" xảy ra <=>
\(\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(y-11\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\y-11=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-11\\y=11\end{matrix}\right.\)
Vậy Min F = 52 <=> x = -11; y = 11
e, \(x^2+12x+y^2-2y+7\)
\(=x^2+12x+36+y^2-2y+1-30\)
\(=\left(x+6\right)^2+\left(y-1\right)^2-30\)
Với mọi giá trị của \(x;y\in R\) ta có: \(\left(x+6\right)^2+\left(y-1\right)^2\ge0\) \(\Rightarrow\left(x+6\right)^2+\left(y-1\right)^2-30\ge-30\) Để \(\left(x+6\right)^2+\left(y-1\right)^2-30=-30\) thì\(\left\{{}\begin{matrix}\left(x+6\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-6\\y=1\end{matrix}\right.\)
Vậy................
