a, \(A=2x^2-8x-10=2\left(x^2-4x+4\right)-18=2\left(x-2\right)^2-18\ge-18\)
Dấu "=" xảy ra <=> x-2=0 <=> x=2
Vậy MinA = -18 khi x=2
b, \(B=x-x^2=-\left(x^2-x+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Dấu "=" xảy ra <=> x-1/2=0 <=> x=1/2
Vậy MaxB = 1/4 khi x=1/2
a) \(A=2x^2-8x-10\)
\(=2\left(x^2-4x-5\right)\)
\(=2\left(x^2-2.x.2+2^2-2^2-5\right)\)
\(=2\left[\left(x-2\right)^2-9\right]\)
\(=2\left(x-2\right)^2-18\)
Vì \(2\left(x-2\right)^2\ge0\forall x\)
Nên \(2\left(x-2\right)^2\ge-18\)
Hay \(A\ge-18\)
Vậy gtnn của A là -18 khi \(2\left(x-2\right)^2=0\)
\(x-2=0\)
\(x=2\)
b) \(B=x-x^2\)
\(=-x^2-x\)
\(=-\left(x^2-x\right)\)
\(=-\text{[}x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\text{]}\)
\(=-\text{[}\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\text{]}\)
\(=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\)
Vì \(-\left(x-\frac{1}{2}\right)^2\le0\forall x\)
Nên \(-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x
\)
Vậy gtln của B là \(\frac{1}{4}\)khi \(x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)