Question

[Trương Võ Thanh Ngân]

lxl + lx-2l=2

Answer 1

[Nguyễn Thị Mai Anh]

Có: \(\left|x\right|+\left|x-2\right|=\left|x\right|+\left|2-x\right|\ge\left|x+2-x\right|=2\)

Để \(\left|x\right|+\left|x-2\right|=2\) thì \(x\left(2-x\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\2-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x\le2\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x\ge2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}0\le x\le2\left(TM\right)\\0\ge x\ge2\left(loai\right)\end{matrix}\right.\)

Vậy \(0\le x\le2\)

Answer 2

[Nguyễn Thị Mai Anh]

phần đầu đúng ko?

* Ta đi CM tổng quát: \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) (1)

Có: \(\left\{{}\begin{matrix}\left|a\right|\ge a;\left|a\right|\ge-a\\\left|b\right|\ge b;\left|b\right|\ge-b\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left|a\right|+\left|b\right|\ge a+b\\\left|a\right|+\left|b\right|\ge-\left(a+b\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left|a\right|+\left|b\right|\ge a+b\\-\left(\left|a\right|+\left|b\right|\right)\le a+b\end{matrix}\right.\)

\(\Rightarrow-\left(\left|a\right|+\left|b\right|\right)\le a+b\le\left|a\right|+\left|b\right|\)

\(\Rightarrow\left|a\right|+\left|b\right|\ge\left|a+b\right|\left(ĐPCM\right)\)

Có: \(\left|x\right|+\left|x-2\right|=\left|x\right|+\left|2-x\right|\) (2)

Áp dụng t/c (1) vào (2), ta đc: \(\left|x\right|+\left|2-x\right|\ge\left|x+2-x\right|=\left|2\right|=2\)