Question

Sa se rezolve in R ecuatia | X-1 |-3|X+4|= X

Answer 1

[tex]|x-1|-3|x+4|= x \\ \\ |x-1|-3|x+4| -x = 0\\ \\ |x-1|-3|x+4|- x = 0 \Leftrightarrow \\ \\ \Leftrightarrow \left\| \begin{array}{ll} -(x-1)+3(x+4) -x = 0, \quad x\in (-\infty,-4)\\ -(x-1)-3(x+4) -x= 0, \quad x\in [-4,1) \\ (x-1)-3(x+4) -x= 0,\quad x\in [1,\infty) \end{array} \right | \Leftrightarrow \\ \\ \\ [/tex]

$\Leftrightarrow \left\| \begin{array}{ll} -x+1+3x+12-x = 0, \quad x\in (-\infty,-4)\\ -x+1-3x-12 -x= 0, \quad x\in [-4,1) \\ x-1-3x-12 -x= 0,\quad x\in [1,\infty) \end{array} \right | \Leftrightarrow \\ \\ \\ \Leftrightarrow \left\| \begin{array}{ll} x=-13 , \quad x\in (-\infty,-4)\\ -5x = 11, \quad x\in [-4,1) \\ -3x = 13,\quad x\in [1,\infty) \end{array} \right | \Leftrightarrow$

$\Leftrightarrow \left\| \begin{array}{ll} x=-13 , \quad x\in (-\infty,-4)\\ \\ x = -\dfrac{11}{5}, \quad x\in [-4,1) \\ \\x = -\dfrac{13}{3},\quad x\in [1,\infty)\quad (F) \end{array} \right | \Rightarrow \boxed{S = \left\{-13,-\dfrac{11}{5}\right\}}$