Question

[Hatake Kakashi]

tìm x biết: a) | 4 - x | + | x + 1 | = 5

                 b) | x + 1 | + | x + 2 | + | x + 3 | + | x + 4 | = 4

                 c) | x - 1 | + | x - 2 | + . . . + | x - 100 |  = 2500

                 d) | x - 1 | + | x - 2 | + | y - 3 | + | x - 4 | = 3

                 e) | x - 4 | + | x -10 | + | x + 101 | + | x + 990 | + | x + 1000 | = 2004

                 f) | x - 2018 | + | x - 2y | \(\le\)0

Answer 1

[Nguyệt]

a)\(\left|4-x\right|+\left|x+1\right|=5\)

\(\left|4-x\right|+\left|x+1\right|\ge\left|4-x+x+1\right|=5\)

dấu = xảy ra khi (4-x).(x+1)=0

=> \(-1\le x\le4\)

b) \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|=4\)

\(\left|x+1\right|+\left|x+4\right|=\left|x+1\right|+\left|-x-4\right|\ge\left|x+1-x-4\right|=\left|-3\right|=3\)

dấu = xảy ra khi \(\left(x+1\right).\left(-x-4\right)\ge0\)

\(-4\le x\le-1\)

\(\left|x+2\right|+\left|x+3\right|=\left|x+2\right|+\left|-x-3\right|\ge\left|x+2-x-3\right|=\left|-1\right|=1\)

dấu = xảy ra khi \(\left(x+2\right).\left(-x-3\right)\ge0\)

\(-3\le x\le-2\)

Answer 2

[Nguyệt]

eiiiiiii sorry nha thiếu, làm tiếp nè =))

\(để\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|=4\)

=> dấu = xảy ra khi đồng thời  \(\hept{\begin{cases}\left|x+1\right|+\left|-x-4\right|=3\\\left|x+2\right|+\left|-x-3\right|=1\end{cases}}\)

Vậy \(-3\le x\le-2\)