Question

Tìm Min:  a,x²+3x+1.                                b, 9x²+3x+1.                Tìm Max : -x²+2x-1 

Answer 1

a/ x² + 3x + 1

\(=x^2+2.\frac{3}{2}.x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+1\)

\(=\left(x+\frac{3}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)

       Vậy MinA = -5/4 khi x + 3/2 = 0 => x = -3/2

b/ 9x² + 3x + 1

\(=\left(3x\right)^2+2.3x.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+1\)

\(=\left(3x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

         Vậy MinB = 3/4 khi 3x + 1/2 = 0 => 3x = -1/2 => x = -1/6

c/ -x² + 2x - 1 = -(x² - 2x + 1) = -(x - 1)² \(\le0\)

          Vậy MaxC = 0 khi x - 1 = 0 => x = 1

Answer 2

a.\(=x^2+2.\frac{3}{2}x+\frac{9}{4}-\frac{5}{4}=\left(x+\frac{3}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)

dấu = xảy ra khi x=-3/2

b,c tt