Question

[dat]

tim gia tri lon nhat

B=\(-3x^2+x+1\)

Answer 1

[hgf]

\(B=-3x^2+x+1\)

\(B=-3\left(x^2-\dfrac{1}{3}x-\dfrac{1}{3}\right)\)

\(B=-3\left[\left(x^2-2\cdot x\cdot\dfrac{1}{6}+\dfrac{1}{36}\right)-\dfrac{13}{36}\right]\)

\(B=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{13}{12}\)\(\le\dfrac{13}{12}\forall x\)

\(B=\dfrac{13}{12}\Leftrightarrow-3\left(x-\dfrac{1}{6}\right)^2=0\Leftrightarrow x=\dfrac{1}{6}\)

Vậy Max B = 13/12 <=> x = 1/6