a: \(=x^2-20x+100+1=\left(x-10\right)^2+1>=1\)
Dấu = xảy ra khi x=1
b: \(=-4\left(x^2-\dfrac{3}{4}x-\dfrac{7}{4}\right)\)
\(=-4\left(x^2-2\cdot x\cdot\dfrac{3}{8}+\dfrac{9}{64}-\dfrac{121}{64}\right)\)
\(=-4\left(x-\dfrac{3}{8}\right)^2+\dfrac{121}{16}< =\dfrac{121}{16}\)
Dấu '=' xảy ra khi x=3/8
a, \(A=x^2-20x+100+1=\left(x-10\right)^2+1\ge1\)
Dấu ''='' xảy ra khi x = 10
b, \(B=-\left(4x^2-3x\right)+7=-\left(4x^2-\dfrac{2.2x.3}{4}+\dfrac{9}{16}-\dfrac{9}{16}\right)+7\)
\(=-\left(2x-\dfrac{3}{4}\right)^2+\dfrac{121}{16}\le\dfrac{121}{16}\)
Dấu ''='' xảy ra khi x = 3/8
\(a.A=x^2-20x+101\)
\(=x^2-20x+100+1\)
\(=\left(x+10\right)^2+1\ge1\)
Vậy \(Min_A=1\) khi \(x+10=0\Leftrightarrow x=-10\)
\(b.B=3x-4x^2+7\)
\(=-\left(4x^2-3x+\dfrac{9}{16}-\dfrac{121}{16}\right)\)
\(=-\left[\left(2x-\dfrac{3}{4}\right)^2-\dfrac{121}{16}\right]\)
\(=-\left(2x-\dfrac{3}{4}\right)^2+\dfrac{121}{16}\le\dfrac{121}{16}\)
Vậy \(Max_B=\dfrac{121}{16}\) khi \(2x-\dfrac{3}{4}=0\Leftrightarrow x=\dfrac{3}{8}\)
