Question

[Trâm Lê Bảo Phương]

Giúp mk vs tìm GTNN của M=x^2-3x+5

Answer 1

[Trang Pham]

M= \(x^2-3x+5=x^2-2\times\frac{3}{2}\times x+\frac{9}{4}-\frac{9}{4}+5\)

M =   \(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)

Vì \(\left(x-\frac{3}{3}\right)^2\ge0\)

=> \(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)

Vậy MIN  M = \(\frac{11}{4}\)dấu bằng xảy ra khi và chỉ khi \(x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)

Answer 2

[Nguyễn Huệ Lam]

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

\(=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)

Vậy \(MinM=\frac{11}{4}\Leftrightarrow\left(x-\frac{3}{2}\right)^2=0\Leftrightarrow x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)