Question

разрешите неравенство |x+2|+|x-1|-|x-4|>3 даю 34бала

Answer 1

$|x+2|+|x-1|-|x-4|>3\\x+2=0; to ; x_1=-2; ; ;; ; x-1=0; to ; x_2=1; ; ;; ; x-4=0; to ; x_3=4\\a); xleq -2:; |x+2|=-(x+2); ;; |x-1|=-(x-1); ;; |x-4|=-(x-4);\\-x-2-x+1+x-4>3; ;; -x>8; to ; underline {x<-8}\\b); -2<xleq 1:; |x+2|=x+2; ;; |x-1|=-(x-1); ;; |x-4|=-(x-4);\\x+2-x+1+x-4>3; ;; x>4; ;\\left { {{-2<xleq 1} atop {x>4}} right. ; ; Rightarrow ; ; underline {xin varnothing }\\c); 1<xleq 4:; |x+2|=x+2; ;; |x-1|=x-1; ;; |x-4|=-(x-4);\\x+2+x-1+x-4>3; ;; 3x>6; ;; x>2; ;$

$left { {{1<xleq 4} atop {x>2}} right. ; ; to ; ; underline {2<xleq 4}\\d); x>4:; |x+2|=x+2; ;; |x-1|=x-1; ;; |x-4|=x-4;\\x+2+x-1-x+4>3; ;; x>-2; ;\\left { {{x>4} atop {x>-2}} right. ; ; to ; ; underline {x>4}\\Otvet:; ; xin (-infty ,-8)cup (2,4, ]cup (4,+infty ); Rightarrow ; xin (-infty ,-8)cup (2,+infty ); .$