\(Q=\sqrt{9x^2-6x+1}+\sqrt{25-30+9x^2}+2011\)
\(Q=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(5-3x\right)^2}+2011\)
\(Q=\left|3x-1\right|+\left|5-3x\right|+2011\)
Đặt \(Q'=\left|3x-1\right|+\left|5-3x\right|\ge\left|3x-1+5-3x\right|=4\)
Đẳng thức xảy ra \(\Leftrightarrow\left(3x-1\right)\left(5-3x\right)\ge0\)
\(\Leftrightarrow\frac{1}{3}\le x\le\frac{5}{3}\)
\(\Rightarrow Min_Q=Min_{Q'}+2011=4+2011=2015\)
Q = \(\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}+2011\)
Q = \(\sqrt{\left(3x-1\right)^2}+\sqrt{\left(3x-5\right)^2}+2011\)
Q = \(3x-1+3x-5+2011\)
Q = \(6x+2005\)
\(Q=\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}+2011\)
\(=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(3x-5\right)^2}+2011\)
\(=\left|3x-1\right|+\left|3x-5\right|+2011\)
Áp dụng BĐT \(\left|x\right|+\left|y\right|\ge\left|x+y\right|\)
\(\left|3x-1\right|+\left|3x-5\right|\ge\left|\left(3x-1\right)+\left(5-3x\right)\right|=4\)
(Dấu "="\(\Leftrightarrow\left(3x-1\right)\left(5-3x\right)\ge0\)
\(TH1:\hept{\begin{cases}3x-1\ge0\\5-3x\ge0\end{cases}}\Leftrightarrow\frac{1}{3}\le x\le\frac{5}{3}\)
\(TH2:\hept{\begin{cases}3x-1\le0\\5-3x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le\frac{1}{3}\\x\ge\frac{3}{5}\end{cases}}\left(L\right)\))
\(\Rightarrow Q\ge2015\)
(Dấu "="\(\Leftrightarrow\frac{1}{3}\le x\le\frac{5}{3}\))
Vậy \(Q_{min}=2015\Leftrightarrow\frac{1}{3}\le x\le\frac{5}{3}\)
