steve jobs
17/03/2017 at 11:00-
We have :
\(\left\{{}\begin{matrix}b< a\\b< 0\\c< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}b-a< 0\\b+c< 0\\c< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left|b-a\right|=a-b\\\left|b+c\right|=-b-c\\\left|c\right|=-c\end{matrix}\right.\)
=> |b - a| + |b + c| + |c| = a - b - b - c - c = a - 2b - 2c
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¤« 08/04/2018 at 15:12
We have :
⎧⎩⎨⎪⎪b<ab<0c<0⇒⎧⎩⎨⎪⎪b−a<0b+c<0c<0
⇒⎧⎩⎨⎪⎪|b−a|=a−b|b+c|=−b−c|c|=−c
=> |b - a| + |b + c| + |c| = a - b - b - c - c = a - 2b - 2c