Ta có:\(B=x^2+8x\)
\(B=x^2+8x+16-16\)
\(B=\left(x+4\right)^2-16\ge-16\)
"="<=>x=-4
\(C=5x^2+x+7\)
\(C=\dfrac{1}{5}\left(25x^2+5x+35\right)\)
\(C=\dfrac{1}{5}\left(25x^2+5x+\dfrac{1}{4}\right)+\dfrac{139}{20}\)
\(C=\dfrac{1}{5}\left(5x+\dfrac{1}{2}\right)^2+\dfrac{174}{25}\ge\dfrac{139}{20}\)
"="<=>x=-0,1
\(D=\dfrac{3}{-4x^2+4x-7}\)
Ta có:\(-4x^2+4x-7=-\left(4x^2-4x+1\right)-6=-\left(2x-1\right)^2-6\le-6\)
\(\Rightarrow D\ge\dfrac{3}{-6}=-\dfrac{1}{2}\)
"="<=>x=0,5
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