Toán Lớp 9: Với x>0 Tim GTNN của A = 4$x^{2}$ -3x + $\frac{1}{4x}$ +2012
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 9: Với x>0 Tim GTNN của A = 4$x^{2}$ -3x + $\frac{1}{4x}$ +2012
Comments ( 2 )
\quad A = 4x^2 -3x + \dfrac{1}{4x} + 2012\\
\to A = \left(x^2 + \dfrac{1}{8x} + \dfrac{1}{8x}\right) + \left(3x^2 – 3x + \dfrac34\right) + 2012 – \dfrac34\\
\to A \geqslant 3\sqrt[3]{x^2\cdot \dfrac{1}{8x}\cdot \dfrac{1}{8x}} + 3\left(x- \dfrac12\right)^2 + 2012 – \dfrac34\\
\to A \geqslant\dfrac34 + 0 +2012 – \dfrac34\\
\to A \geqslant 2012\\
\text{Dấu = xảy ra khi và chỉ khi}\\
\begin{cases}x^2 = \dfrac{1}{8x}\\x – \dfrac12 = 0\end{cases}\Leftrightarrow x = \dfrac12\\
\text{Vậy}\ \min A = 2012\Leftrightarrow x = \dfrac12
\end{array}\)