\(-4x^2+5x-21\)
\(=-4\left(x^2-\frac{5}{4}x\right)-21\)
\(=-4\left(x^2-2x.\frac{5}{8}+\frac{25}{64}\right)-21+\frac{25}{16}\)
\(=-4\left(x-\frac{5}{8}\right)^2-\frac{311}{16}\)
Có \(\left(x-\frac{5}{8}\right)^2\ge0\) với mọi x
=> \(-4\left(x-\frac{5}{8}\right)^2\le0\)với mọi x
=> \(-4\left(x-\frac{5}{8}\right)^2-\frac{311}{16}\le\frac{-311}{16}\)với mọi x
Dấu "=" xảy ra <=> \(x-\frac{5}{8}=0\)<=> \(x=\frac{5}{8}\)
KL: GTLN của biểu thức là \(\frac{-311}{16}\)<=> \(x=\frac{5}{8}\)
\(A=-4x^2+5x-21\)
\(=-\left[\left(2x\right)^2-2\times2x\times\frac{5}{4}+\left(\frac{5}{4}\right)^2-\left(\frac{5}{4}\right)^2+21\right]\)
\(=-\left[\left(2x-\frac{5}{4}\right)^2+\frac{311}{16}\right]\)
\(\left(2x-\frac{5}{4}\right)^2\ge0\)
\(\left(2x-\frac{5}{4}\right)^2+\frac{311}{16}\ge\frac{311}{16}\)
\(-\left[\left(2x-\frac{5}{4}\right)^2+\frac{311}{16}\right]\le-\frac{311}{16}\)
Vậy Max A = \(-\frac{311}{16}\) khi x = \(\frac{5}{8}\)
