Question

I 5-I 2-3xII=x-1 тема модулі

Answer 1

$|5-|2-3x|, |=x-1quad Rightarrow \\5-|2-3x|=pm (x-1)\\a); ; 5-|2-3x|=-(x-1)\\|2-3x|=5+(x-1)$

$|2-3x|=x+4; ; to ; ; left [ {{2-3x=x+4,; x leq frac{2}{3}} atop {2-3x=-(x+4),; x>frac{2}{3}}} right. ; left [ {{4x=-2,; x leq frac{2}{3}} atop {2x=6,; x>frac{2}{3}}} right.$
 $left [ {{4x=-2,; x leq } atop {2x=6,; x>frac{2}{3}}} right. left [ {{x=-frac{1}{2}} atop {x=3}} right. \\b); ; 5-|2-3x|=x-1\\|2-3x|=6-x; ; to ; ; left [ {{2-3x=6-x,; x leq frac{2}{3}} atop {2-3x=-(6-x),; x>frac{2}{3}}} right. ; left [ {{2x=-4} atop {4x=8}} right. ; left { {{x=-2} atop {x=2}} right.$

$b); ; 5-|2-3x|=x-1\\|2-3x|=6-x; ; to ; ; left [ {{2-3x=6-x,; x leq frac{2}{3}} atop {2-3x=-(6-x),; x>frac{2}{3}}} right. ; left [ {{2x=-4} atop {4x=8}} right. ; left { {{x=-2} atop {x=2}} right. \\Proverka:; 1); x=-frac{1}{2}:; |5-|2+3cdot frac{1}{2}|, |=|5-frac{7}{2}|=frac{3}{2}; ;\\x-1=-frac{1}{2}-1=-frac{1}{2}-1=-frac{3}{2}; ;\\frac{3}{2}ne -frac{3}{2}\\2); ; x=3:; ; |5-|2-3cdot 3||=|5-7|=|-2|=2; ;\\x-1=3-1=2; ;\\2=2$

$3); ; x=-2:; ; |5-|2+3cdot 2||=|5-8|=|-3|=3; ;\\x-1=-2-1=-3; ;\\3ne -3\\4); ; x=2:; ; |5-|2-3cdot 2||=|5-4|=1\\x-1=2-1=1\\1=1\\Otvet:; ; x=3; ,; ; x=2; .$