\(\left|4x+3\right|-x=5\)
\(\Leftrightarrow\left|4x+3\right|=5+x\)
+ ) Khi \(4x+3\ge0\Leftrightarrow x\ge-\dfrac{3}{4}\)
\(4x+3=5+x\)
\(\Leftrightarrow3x=2\)
\(\Leftrightarrow x=\dfrac{2}{3}\) ( chọn )
+ ) Khi \(4x+3< 0\Leftrightarrow x< -\dfrac{3}{4}\)
\(-4x-3=5+x\)
\(\Leftrightarrow-5x=8\)
\(\Leftrightarrow x=-\dfrac{8}{5}\) ( chọn )
Vậy ........
Ta có : \(\left|4x+3\right|-x=5\)
\(\Leftrightarrow\left|4x+3\right|=5+x\)
Trường hợp 1 : \(4x+3\ge0\Leftrightarrow x< -\dfrac{3}{4}\)
\(\Leftrightarrow4x+3=5+x\)
\(\Leftrightarrow3x=2\)
\(\Leftrightarrow x=\dfrac{2}{3}\) ( Thỏa mãn )
Trường hợp 2 :\(4x+3< 0\Leftrightarrow x< -\dfrac{3}{4}\)
\(-4x-3=5+x\)
\(\Leftrightarrow-5x=8\)
\(\Leftrightarrow x=\dfrac{-8}{5}\) ( Thỏa mãn )
Vậy ...
