\(A=9x^2+18x-20\)
\(\Leftrightarrow A=\left(3x\right)^2+2.3x.3+9-29\)
\(\Leftrightarrow A=\left(3x+3\right)^2-29\le-29\forall x\)
Dấu " = " xảy ra
\(\Leftrightarrow\left(3x+3\right)^2=0\Leftrightarrow3x+3=0\Leftrightarrow3x=-3\Leftrightarrow x=-1\)
Vậy Min A là : \(-29\Leftrightarrow x=-1\)
\(B=m^2+10m+1\)
\(\Leftrightarrow B=m^2+2.m.5+25-24\)
\(\Leftrightarrow B=\left(m+5\right)^2-24\le-24\forall m\)
Dấu \("="\) xảy ra
\(\Leftrightarrow\left(m+5\right)^2=0\Leftrightarrow m+5=0\Leftrightarrow m=-5\)
Vậy Min B là : -24 \(\Leftrightarrow m=-5\)
\(C=25x^2-20x+30\)
\(\Leftrightarrow C=\left(5x\right)^2-2.5x.2+4+26\)
\(\Leftrightarrow C=\left(5x-2\right)^2+26\le26\forall x\)
Dấu " = " xảy ra
\(\Leftrightarrow\left(5x-2\right)^2=0\Leftrightarrow5x-2=0\Leftrightarrow5x=2\Leftrightarrow x=\dfrac{2}{5}\)
Vậy Min C là : 26 \(\Leftrightarrow x=\dfrac{2}{5}\)
Sorry , sửa lại :
\(\le\) \(\Rightarrow\) \(\ge\)