\(A=\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}\)
\(=\sqrt{9x^2-6x+1}+\sqrt{9x^2-30x+25}\)
\(=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(3x-5\right)^2}\)
\(=\left|3x-1\right|+\left|3x-5\right|\)
\(=\left|3x-1\right|+\left|5-3x\right|\)
\(\ge\left|3x-1+5-3x\right|=4\)
Xảy ra khi \(\dfrac{1}{3}\le x\le\dfrac{5}{3}\)
\(A=\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}\)
\(=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(3x-5\right)^2}\)
\(=\left|3x-1\right|+\left|3x-5\right|=\left|3x-1\right|+\left|5-3x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:
\(A\ge\left|3x-1+5-3x\right|=\left|4\right|=4\)
Dấu " = " khi \(\left\{{}\begin{matrix}3x-1\ge0\\5-3x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\x\le\dfrac{5}{3}\end{matrix}\right.\)
Vậy \(MIN_A=4\) khi \(\dfrac{1}{3}\le x\le\dfrac{5}{3}\)