Tham khảo nhé :
\(\left|x-2\right|-3x\left|+\right|x-1\left|x-3\right|=0\)
\(\Leftrightarrow\left|x-2\right|-3=0\)hoặc \(5+\left|x\right|=0\)
Xét \(\left|x-2\right|-3=0\Leftrightarrow\left|x-2\right|=3\)
\(\Rightarrow x-2=\pm3\)
Với: \(x-2=3\Rightarrow x=5\)
Với: \(x-2=-3\Rightarrow x=-1\)
Vậy:..
x2-3x >0 (trong gia tri tuyet doi nen >0)
(x+1)(x-3) >0 (trong gia tri tuyet doi nen > 0)
\( \left|x^2-3x\right|+\left|\left(x+1\right)\left(x-3\right)\right|=0\)
\(\Leftrightarrow\left|x\left(x-3\right)\right|+\left|\left(x+1\right)\left(x-3\right)\right|=0\)
Với \(x< -1\), ta có \(pt\Leftrightarrow x\left(x-3\right)+\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=3\\x=-\frac{1}{2}\end{cases}\left(l\right)}\)
Với \(-1\le x< 0\)ta có \(pt\Leftrightarrow x\left(x-3\right)-\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(l\right)\)
Với \(0\le x< 3\)ta có \(pt\Leftrightarrow-x\left(x-3\right)-\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-2x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=3\\x=-\frac{1}{2}\end{cases}}\left(l\right)\)
Với \(x\ge3\)ta có \(pt\Leftrightarrow x\left(x-3\right)+\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=3\left(n\right)\\x=-\frac{1}{2}\left(l\right)\end{cases}}\)
Vậy x = 3.