Question

Tìm GTLN:

\(A=4x-4x^2\)

\(B=\dfrac{3}{\left|2x-5\right|+2}\)

\(C=\dfrac{5}{x^2-3x+4}\)

Answer 1

*\(A=4x-4x^2=-\left(2x\right)^2+4x=-\left[\left(2x\right)^2-2\cdot2x\cdot1+1^2\right]+1=-\left(2x-1\right)^2+1\le1\)

Vậy \(A_{max}=1\Leftrightarrow x=\dfrac{1}{2}\)

* Ta có:

\(\left|2x-5\right|\ge0\Rightarrow\left|2x-5\right|+2\ge2\)

\(\Rightarrow B\le\dfrac{3}{2}\)

Vậy \(B_{max}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{5}{2}\)

* Ta có:

\(x^2-3x+4=\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}\right)+\dfrac{7}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)

\(\Rightarrow C\le\dfrac{5}{\dfrac{7}{4}}=\dfrac{20}{7}\)

Vậy \(C_{max}=\dfrac{20}{7}\Leftrightarrow x=\dfrac{3}{2}\)