e, \(E=\sqrt{x^2-2x+1}+\sqrt{x^2-6x+9}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-3\right)^2}\)
\(=\left|x-1\right|+\left|x-3\right|=\left|x-1\right|+\left|3-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:
\(E\ge\left|x-1+3-x\right|=\left|2\right|=2\)
Dấu " = " khi \(\left\{{}\begin{matrix}x-1\ge0\\3-x\ge0\end{matrix}\right.\Rightarrow1\le x\le3\)
Vậy \(MIN_E=2\) khi \(1\le x\le3\)
f, \(F=\sqrt{x+9-6\sqrt{x}}+\sqrt{x+1-2\sqrt{x}}\)
\(=\sqrt{\left(\sqrt{x}-3\right)^2}+\sqrt{\left(\sqrt{x}-1\right)^2}\)
\(=\left|\sqrt{x}-3\right|+\left|\sqrt{x}-1\right|=\left|3-\sqrt{x}\right|+\left|\sqrt{x}-1\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:
\(F\ge\left|3-\sqrt{x}+\sqrt{x}-1\right|=\left|2\right|=2\)
Dấu " = " khi \(\left\{{}\begin{matrix}3-\sqrt{x}\ge0\\\sqrt{x}-1\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\le\sqrt{3}\\x\ge1\end{matrix}\right.\)
Vậy \(MIN_F=2\) khi \(1\le x\le\sqrt{3}\)
