Question

tìm GTLN

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

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

c. C= -3x⁴ +4x -1

Answer 1

Làm 1 câu thôi các câu còn lại tự làm :v

B đạt MAX \(\Leftrightarrow3x^2-2x+5\) đạt MIN .

\(3x^2-2x+5=3\left(x^2-\dfrac{2}{3}x+\dfrac{5}{3}\right)=3\left[\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)+\dfrac{14}{9}\right]=3\left[\left(x-\dfrac{1}{3}\right)^2+\dfrac{14}{9}\right]\ge\dfrac{14}{3}\)

\(\Rightarrow B\le\dfrac{2}{\dfrac{14}{3}}=\dfrac{3}{7}\) Dấu \(''=''\) xảy ra khi \(\left(x-\dfrac{1}{3}\right)^2=0\Leftrightarrow x=\dfrac{1}{3}\)

Answer 2

 

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