Question

Помогите решить 2Б.188.

Answer 1

$|4-x|+|x-2|=2\\4-x=0; ; to ; ; x=4; ,; ; znaki; (4-x):; ; +++(4)---\\x-2=0; ; to ; ; x=2; ,; ; znaki; (x-2):; ; ---(2)+++\\a); ; xleq 2:; ; |4-x|=+(4-x); ,; |x-2|=-(x-2); ,\\|4-x|+|x-2|=4-x-(x-2)=-2x+6; ,; ; \\-2x+6=2; ,; 2x=4; ,; ; underline {x=2}in (-infty ,2, ]\\b); ; 2<xleq 4:; ; |4-x|=+(4-x); ,; |x-2|=+(x-2); ,\\|4-x|+|x-2|=4-x+x-2=2; ,; ; 2=2; ; to \\underline {xin (2,4, ]}\\c); ; x>4:; ; |4-x|=-(4-x); ,; |x-2|=+(x-2); ,\\|4-x|+|x-2|=-4+x+x-2=2x-6; ,\\2x-6=2; ,; 2x=8; ,; x=4notin (4,+infty )\\Otvet:; ; xin {2}cup (2,4, ]=[, 2,4, ]; .$