\(A=-2x^2+8x-3=-\left(2x^2-8x+3\right)=-\left[2.\left(x^2-4x+\frac{3}{2}\right)\right]\)
\(=-\left[2.\left(x^2-2.x.2+2^2-2^2+\frac{3}{2}\right)\right]=-\left[2.\left(\left(x-2\right)^2-\frac{5}{2}\right)\right]=-\left[2\left(x-2\right)^2-5\right]\)
\(=5-2\left(x-2\right)^2\le5\) với mọi x
=>minA=5
Dấu "=" xảy ra <=> x=2
Vậy.................