\(A=x^2+10x-37=x^2+2x.5+25-62\)\(=\left(x+5\right)^2-62\)
Vì \(\left(x+5\right)^2\ge0\Rightarrow A\ge-62\)
Dấu ''='' xảy ra \(\Leftrightarrow x=-5\)
Vậy ...
\(C=6x-x^2+3=-\left(x^2-2x.3+9\right)+12=\)\(-\left(x-3\right)^2+12\)
Vì \(-\left(x-3\right)^2\le0\Rightarrow C\le12\)
Dấu ''='' xảy ra khi x =3
Vậy ....
Câu b,d làm tương tự nhé :)
\(B=4\left(x^2-\dfrac{3}{4}x+\dfrac{1}{4}\right)\)
\(=4\left(x^2-2\cdot x\cdot\dfrac{3}{8}+\dfrac{9}{64}+\dfrac{7}{64}\right)\)
\(=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{7}{16}>=\dfrac{7}{16}\)
Dấu '=' xảy ra khi x=3/8
\(D=-\left(x^2-2x+3\right)\)
\(=-\left(x^2-2x+1+2\right)\)
\(=-\left(x-1\right)^2-2< =-2\)
Dấu '=' xảy ra khi x=1
