Question

tim GTNN cua cac bieu thuc

a)2x\(^2\)+8x+1

b)2x\(^2\)+3x+1

Answer 1

a/ \(2x^2+8x+1=2\left(x^2+4x+\frac{1}{2}\right)=2\left(x^2+2.2x+4-4+\frac{1}{2}\right)\)

     \(=2\left[\left(x+2\right)^2-\frac{7}{2}\right]=2\left(x+2\right)^2-7\ge-7\)

       Vậy Min A = -7 khi x + 2 = 0 => x = 2

b/ \(2x^2+3x+1=2\left(x^2+\frac{3}{2}x+\frac{1}{2}\right)=2\left(x^2+2.\frac{3}{4}.x+\frac{9}{16}-\frac{9}{16}+\frac{1}{2}\right)\)

          \(=2\left[\left(x+\frac{3}{4}\right)^2-\frac{1}{16}\right]=2\left(x+\frac{3}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\)

         Vậy Min B = -1/8 khi x + 3/4 = 0 => x = -3/4