Question

СРОЧНО РЕШИТЕ ХОТЯБЫ ОДНО

Answer 1

$1); ; | x^2-16x+36|leq |36-x^2|\\star ; ; |x|<a; ;Leftrightarrow ; ; -a<x<a; ; star \\-(36-x^2)leq x^2-16x+36leq 36-x^2; ; Leftrightarrow ; ; ; left { {{x^2-36leq x^2-16x+36} atop {x^2-16x+36leq 36-x^2}} right. \\left { {{16x-72leq 0} atop {2x^2-16xleq 0}} right. ; left { {{xleq 4,5} atop {x(x-8)leq 0}} right. ; left { {{xleq 4,5} atop {0leq xleq 8}} right. ; ; Rightarrow ; ; underline {xin [, 0, ;, 4,5, ]}; -; ; otvet.\\star ; x(x-8)leq 0; ,; ; znaki, :; ; ; +++[, 0, ]---[, 8, ]+++; ; ,; ; xin [, 0,8, ]$

$2); ; Big |frac{x^2-5x+4}{x^2-4}Big |leq 1; ; Leftrightarrow ; ; -1leq frac{x^2-5x+4}{x^2-4}leq 1\\left { {{frac{x^2-5x+4}{x^2-4}-1leq 0} atop {frac{x^2-5x+4}{x^2-4}+1geq 0} right. ; ; left { {{frac{x^2-5x+4-x^2+4}{x^2-4}leq 0} atop {frac{x^2-5x+4+x^2-4}{x^2-4}geq 0}} right. ; left { {{frac{-(5x-8)}{(x-2)(x+2)}leq 0} atop {frac{2x^2-5x}{x^2-4}geq 0}} right. ; left { {{frac{5x-8}{(x-2)(x+2)}geq 0} atop {frac{x(2x-5)}{(x-2)(x+2)}geq 0}} right.$

$a); ; frac{5x-8}{(x-2)(x+2)}geq 0:; ; ---(-2)+++[, 1,6, ]---(2)+++\\xin (-2;, 1,6, ]cup (2,+infty )\\b); ; frac{x(2x-5)}{(x-2)(x+2)}geq 0:; ; +++(-2)---[, 0, ]+++(2)---[2,5, ]+++\\xin (-infty ;-2)cup[, 0;2)cup [, 2,5;+infty )\\Otvet:; ; xin [, 0, ;, 1,6, ]cup [, 2,5, ;+infty ); .$