Question

Решите систему неравенств
1)$left { {{x^2-x+6 textgreater 0} atop {x^2leq25 }} right.$
2)$frac{x^2(1-x)}{x^2-6x-9} leq 0$

Answer 1

$1); ; left { {{x^2-x+6>0} atop {x^2leq 25}} right.; left { {{xin (-infty ,+infty )} atop {(x-5)(x+5)leq 0}} right. ; left { {{xin (-infty ,+infty )} atop {xin [-5.5, ]}} right. ; ; Rightarrow ; ; xin [-5,5; ]\\2); ; frac{x^2(1-x)}{x^2-6x-9}leq 0\\x^2-6x-9=0; ,; D/4=9+9=18; ,; ; x_{1,2}=3pm 3sqrt2\\frac{x^2(x-1)}{(x-3-3sqrt2)(x+3-3sqrt2)} geq 0\\znaki:; ; ; ---(3-3sqrt2)+++[, 0; ]+++[; 1; ]---(3+3sqrt2)+++\\xin (3-3sqrt2, ;, 1; ]cup (3+3sqrt2, ;, +infty , )$