a)Áp dụng BĐT \(|a|+|b|\ge |a+b|\) ta có:
\(A=\left|x-1\right|+\left|x-2017\right|\)
\(=\left|x-1\right|+\left|2017-x\right|\)
\(\ge\left|x-1+2017-x\right|=2016\)
Khi \(1\le x\le 2017\)
b)Ta thấy: \(\left(2x-1\right)^2\ge0\)
\(\Rightarrow B=\left(2x-1\right)+5\ge5\)
Khi \(x=\dfrac{1}{2}\)
Đặt:
\(PHUCDZ=\left|x-1\right|+\left|x-2017\right|\)
\(PHUCDZ=\left|x-1\right|+\left|2017-x\right|\)
Áp dụng bđt: \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
\(PHUCDZ\ge\left|x-1+2017-x\right|\)
\(PHUCDZ\ge2016\)
Dấu "=" xảy ra khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1\ge0\Rightarrow x\ge1\\2017-x\ge0\Rightarrow x\le2017\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\Rightarrow x< 1\\2017-x< 0\Rightarrow x>2017\end{matrix}\right.\end{matrix}\right.\)
Vậy \(1\le x\le2017\)
Đặt:
\(max=\left(2x-1\right)^2+5\)
\(\left(2x-1\right)^2\ge0\forall x\)
\(max=\left(2x-1\right)^2+5\ge5\)
Dấu "=" xảy ra khi:
\(2x-1=0\Rightarrow x=\dfrac{1}{2}\)
Vậy \(min_{max}=5\)
