Question

Tìm Max hoặc Min:

A=3x-x²+4

B=2x²-3x+2.

Answer 1

\(A=-\left(x^2-3x-4\right)\)

\(=-\left(x^2-2.x\frac{3}{2}+\frac{9}{4}+\frac{7}{4}\right)\)

\(=-\left(\left(x-\frac{3}{2}\right)+\frac{7}{4}\right)\)

\(=-\frac{7}{4}-\left(x-\frac{3}{2}\right)^2\le\frac{-7}{4}\)

Vậy \(MAXA=\frac{-7}{4}\Leftrightarrow x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)

Answer 2

\(B=2\left(x^2-\frac{3}{2}x+1\right)=2\left(x^2-2\times x\times\frac{3}{4}+\frac{9}{16}-\frac{9}{16}+1\right)=2\left(x-\frac{3}{4}\right)^2+\frac{7}{8}\ge\frac{7}{8}\)

MIN B = 7/8 <=> x=3/4