Question

[Việt Anh]

Tim GTNN A= x^2-5x+1

GTLN B=1-x^2+3x

Answer 1

[Đặng Tiến]

\(A=x^2-5x+1=x^2-2.x.\frac{5}{2}+\left(\frac{5}{2}\right)^2-\frac{21}{4}=\left(x-\frac{5}{2}\right)^2-\frac{21}{4}\)

Vì \(\left(x-\frac{5}{2}\right)^2\ge0\)

nên \(\left(x-\frac{5}{2}\right)^2-\frac{21}{4}\ge-\frac{21}{4}\)

Vậy \(Min_{x^2-5x+1}=-\frac{21}{4}\)khi \(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\).

\(B=1-x^2+3x=-\left(x^2-3x-1\right)=-\left[x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\frac{13}{4}\right]=-\left[\left(x-\frac{3}{2}\right)^2-\frac{13}{4}\right]=-\left(x-\frac{3}{2}\right)^2+\frac{13}{4}\)Vì \(\left(x-\frac{3}{2}\right)^2\ge0\)

nên \(-\left(x-\frac{3}{2}\right)^2\le0\)

do đó \(-\left(x-\frac{3}{2}\right)^2+\frac{13}{4}\le\frac{13}{4}\)

Vậy \(Max_{1-x^2+3x}=\frac{13}{4}\)khi \(x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)