\(A=x^2-2x+4\)
\(A=\left(x^2-2x+1\right)+3\)
\(A=\left(x-1\right)^2+3\)
Vì \(\left(x-1\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x-1\right)^2+3\ge3\) với mọi x
\(\Rightarrow Amin=3\Leftrightarrow x=1\)
\(B=4x^2-4x+1\)
\(B=\left(2x-1\right)^2\)
Vì \(\left(2x-1\right)^2\ge0\) với mọi x
\(\Rightarrow Bmin=0\Leftrightarrow x=\dfrac{1}{2}\)
\(C=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(C=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(C=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(C=\left(x^2+5x\right)^2-36\)
Vì \(\left(x^2+5x\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x^2+5x\right)^2-36\ge-36\)
\(\Rightarrow Cmin=-36\Leftrightarrow x^2+5x=0\)
\(\Rightarrow x\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
D=x^2-4x+4-3
=(x-2)^2-3>=-3
Dấu = xảy ra khi x=2
x^2-4x+10=(x-2)^2+6>=6
=>4/x^2-4x+10<=2/3
=>D>=-2/3
Dấu = xảy ra khi x=2
x^2-x+1=(x-1/2)^2+3/4>=3/4
=>2/x^2-x+1<=2:3/4=2*4/3=8/3
=>F>=-8/3
dấu = xảy ra khi x=1/2
