Với mọi x ta có :
\(\hept{\begin{cases}\left|x+19\right|\ge0\\\left|x+5\right|\ge0\\\left|x+2011\right|\ge0\end{cases}}\)
\(\Leftrightarrow\left|x+19\right|+\left|x+5\right|+\left|x+2011\right|\ge0\)
Mà \(\left|x+19\right|+\left|x+5\right|+\left|x+2011\right|=4x\)
\(\Leftrightarrow4x\ge0\Leftrightarrow x\ge0\)
Với \(x\ge0\) ta dc :
+) \(\left|x+19\right|=x+19\)
+) \(\left|x+5\right|=x+5\)
\(\left|x+2011\right|=x+2011\)
\(\Leftrightarrow x+19+x+5+x+2011=4x\)
\(\Leftrightarrow3x+2035=4x\)
\(\Leftrightarrow x=2035\)
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