\(C=\sqrt{25x^2-20x+4}+\sqrt{25x^2}\)
\(C=\sqrt{\left(5x-2\right)^2}+\sqrt{\left(5x\right)^2}\)
\(C=\left|5x-2\right|+\left|5x\right|=\left|2-5x\right|+\left|5x\right|\)
\(C\ge\left|2-5x+5x\right|=2\)
Dấu " = " xảy ra \(\Leftrightarrow\)( 2 - 5x ) . 5x \(\ge\)0
\(\Leftrightarrow\)\(\hept{\begin{cases}x\ge0\\2-5x\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x\le0\\2-5x\le0\end{cases}}\)
\(\Leftrightarrow\)\(0\le x\le\frac{2}{5}\)
Vậy GTNN của C là 2 \(\Leftrightarrow\)\(0\le x\le\frac{2}{5}\)
\(C=\sqrt{25x^2-20x+4}+\sqrt{25x^2}\)
\(C=\sqrt{\left(5x-2\right)^2}+\sqrt{\left(5x\right)^2}\)
\(C=\left|5x-2\right|+\left|5x\right|\)
\(C=\left|2-5x\right|+\left|5x\right|\ge\left|2-5x+5x\right|=2\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2-5x\ge0\\5x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le\frac{2}{5}\\x\ge0\end{cases}\Leftrightarrow0\le}x\le\frac{2}{5}}\)
