a) cách1: \(\left|x+1\right|+3=3x\)
th1: \(x+1\ge0\Leftrightarrow x\ge-1\)
\(\Rightarrow\left|x+1\right|+3=3x\Leftrightarrow x+1+3=3x\Leftrightarrow2x=4\Leftrightarrow x=2\left(tmđk\right)\)
th2: \(x+1< 0\Leftrightarrow x< -1\)
\(\Rightarrow\left|x+1\right|+3=3x\Leftrightarrow-x-1+3=3x\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\left(loại\right)\)
vậy \(x=2\)
cách 2: điều kiện : \(3x-3\ge0\Leftrightarrow3x\ge3\Leftrightarrow x\ge1\) \(\left|x+1\right|+3=3x\Leftrightarrow\left|x+1\right|=3x-3\Leftrightarrow\left(\left|x+1\right|\right)^2=\left(3x-3\right)^2\)
\(\Leftrightarrow\left(x+1\right)^2=\left(3x-3\right)^2\Leftrightarrow x^2+2x+1=9x^2-18x+9\)
\(\Leftrightarrow9x^2-18x+9-x^2-2x-1=0\Leftrightarrow8x^2-20x+8=0\)
\(\Leftrightarrow8x^2-4x-16x+8=0\Leftrightarrow4x\left(2x-1\right)-8\left(2x-1\right)=0\)
\(\Leftrightarrow\left(4x-8\right)\left(2x-1\right)=0\Leftrightarrow\left\{{}\begin{matrix}4x-8=0\\2x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x=8\\2x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\left(tmđk\right)\\x=\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)
vậy \(x=2\)
b) cách1: \(\left|x+3\right|-x=4\)
th1: \(x+3\ge0\Leftrightarrow x\ge-3\)
\(\Rightarrow\left|x+3\right|-x=4\Leftrightarrow x+3-x=4\Leftrightarrow3=4\left(vôlí\right)\)
th2: \(x+3< 0\Leftrightarrow x< -3\)
\(\Rightarrow\left|x+3\right|-x=4\Leftrightarrow-x-3-x=4\Leftrightarrow2x=-7\Leftrightarrow x=\dfrac{-7}{2}\left(tmđk\right)\)
vậy \(x=\dfrac{-7}{2}\)
cách2: điều kiện \(4+x\ge0\Leftrightarrow x\ge-4\)
\(\left|x+3\right|-x=4\Leftrightarrow\left|x+3\right|=4+x\Leftrightarrow\left(\left|x+3\right|\right)^2=\left(4+x\right)^2\)
\(\Leftrightarrow\left(x+3\right)^2=\left(4+x\right)^2\Leftrightarrow x^2+6x+9=16+8x+x^2\)
\(\Leftrightarrow x^2+6x+9-16-8x-x^2\Leftrightarrow-2x-7=0\Leftrightarrow-2x=7\Leftrightarrow x=\dfrac{-7}{2}\left(tmđk\right)\)
vậy \(x=\dfrac{-7}{2}\)
