â/ Với \(\forall x\) ta có :
\(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left|x+2\right|\ge0\\\left|x+3\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|\ge0\)
\(Mà\) \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|=4x\)
\(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x\ge0\)
Với \(\forall x\ge0\) ta có :
+) \(\left|x+1\right|=x+1\)
+) \(\left|x+2\right|=x+2\)
+) \(\left|x+3\right|=x+3\)
\(\Leftrightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|=x+1+x+2+x+3=4x\)
\(\Leftrightarrow3x+6=4x\)
\(\Leftrightarrow x=6\)
Vậy ...
Đúng 0
Bình luận (0)
