Question

tìm gt lớn nhất của :

a. B=\(\dfrac{2}{3x^2-2x+5}\)

b. A=\(\dfrac{-2x^2+3x-1}{4}\)

c. C=-3x⁴+4x-1

Answer 1

a: \(3x^2-2x+5\)
\(=3\left(x^2-\dfrac{2}{3}x+\dfrac{5}{3}\right)\)

\(=3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}+\dfrac{14}{9}\right)\)

\(=3\left(x-\dfrac{1}{3}\right)^2+\dfrac{14}{3}>=\dfrac{14}{3}\)

=>B<=2:14/3=2x3/14=6/14=3/7

Dấu '=' xảy ra khi x=1/3

b: \(-2x^2+3x-1\)

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

\(=-2\left(x^2-2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{1}{16}\right)\)

\(=-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}< =\dfrac{1}{8}\)

=>A<=1/32

Dấu = xảy ra khi x=3/4