Question

[Vu Ngoc Thuy Linh_2003]

TINH GIA TRI LON NHAT CUA BIEU THUC

B=9x-3x²

Answer 1

[OoO Pipy OoO]

\(B=9x-3x^2=-3\times\left(x^2-2\times x\times\frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\right)=-3\times\left[\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\right]\)

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

\(\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\)

\(-3\times\left[\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\right]\le\frac{27}{4}\)

Vậy Max B = \(\frac{27}{4}\) khi x = \(\frac{3}{2}\)

Answer 2

[o0o I am a studious person o0o]

\(B=9x-3x^2\)

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

\(=3\left(x^2-2x+1-1\right)\)

\(=-3+3\left(x-1\right)^2\ge-3\)

Max \(B=-3\Leftrightarrow x-1=0\Rightarrow x=1\)