Question

[FFPUBGAOVCFLOL]

Tìm GTNN của \(B=\left|x-1\right|+\left|x-2\right|+...+\left|x-100\right|\)

Answer 1

[Kiệt Nguyễn]

Áp dụng BĐT \(\left|x\right|+\left|y\right|\ge\left|x+y\right|\):

\(\left|x-1\right|+\left|x-100\right|\ge\left|\left(x-1\right)+\left(100-x\right)\right|=99\)

(Dấu "=" khi \(1\le x\le100\))

\(\left|x-2\right|+\left|x-99\right|\ge\left|\left(x-2\right)+\left(99-x\right)\right|=97\)

(Dấu "=" khi \(2\le x\le99\))

\(\left|x-3\right|+\left|x-98\right|\ge\left|\left(x-3\right)+\left(98-x\right)\right|=95\)

(Dấu "=" khi \(3\le x\le98\))

...

\(\left|x-49\right|+\left|x-50\right|\ge\left|\left(x-49\right)+\left(50-x\right)\right|=1\)

(Dấu "="\(\Leftrightarrow49\le x\le50\))

Vậy \(B\ge99+97+95+...+1=\frac{\left(99+1\right)\left[\left(99-1\right):2+1\right]}{2}\)

\(=2500\)

Dấu "=" khi \(49\le x\le50\)