Question

Tìm GTNN của biểu thức sau:

\(B=\frac{3}{2}x^2+x+1\)

Answer 1

B=\(\frac{3}{2}\)x²+x+1

=\(\frac{3}{2}\)(x²+\(\frac{2}{3}\)x)+1

=\(\frac{3}{2}\)(x²+2.\(\frac{1}{3}\)x+\(\frac{1}{9}\))+1-\(\frac{3}{2}.\frac{1}{9}\)

=\(\frac{3}{2}\left(x+\frac{1}{3}\right)^2+\frac{5}{6}\ge\frac{5}{6}\)

Dấu = xảy ra khi : \(\frac{3}{2}\left(x+\frac{1}{3^{ }}\right)^2=0\)

\(\Rightarrow x=-\frac{1}{3}\)

Vậy Min B=\(\frac{5}{6}\)\(\Leftrightarrow x=-\frac{1}{3}\)