Question

Решите неравенство
|x²-4x+3|≤x-1

Answer 1

$|x^{2}-4x+3|\leq x-1$

$-(x-1)\leq x^{2}-4x+3\leq x-1$

$\left \{ {{x^2-4x+3\leq x-1} \atop {x^2-4x+3\geq -x+1}} \right.$

$\left \{ {{x^2-4x+3-x+1\leq 0} \atop {x^2-4x+3+x-1\geq 0}} \right.$

$\left \{ {{x^2-5x+4\leq 0} \atop {x^2-3x+2\geq 0}} \right.$

$\left \{ {{1\leq x\leq 4} \atop {x\leq 1;x\geq 2}} \right.$

Общее решение:

$x$ ∈ {$1$}∪[2; 4]