Question

Tìm giá trị lớn nhất của biểu thức:

\(A=\dfrac{9x^2-12x+14}{\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)...\left(\dfrac{1}{100}-1\right)}\)

Answer 1

Lời giải:

Ta có:

\(M=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).....\left(\frac{1}{100}-1\right)\)

\(-M=M(-1)^{99}=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{100}\right)\)

\(-M=\frac{(2-1)(3-1)...(100-1)}{2.3.4....100}=\frac{1.2.3....99}{2.3.4...100}=\frac{1}{100}\)

\(\Rightarrow M=-\frac{1}{100}\Rightarrow A=-100(9x^2-12x+14)\)

\(\Leftrightarrow A=-100[(3x-2)^2+10]\)

Ta có \((3x-2)^2\geq 0\forall x\in\mathbb{R}\Rightarrow (3x-2)^2+10\geq 10\)

\(\Rightarrow -100[(3x-2)^2+10]\leq -1000\)

Hay \(A_{\max}=-1000\Leftrightarrow x=\frac{2}{3}\)