Question

1) 1/х+1+2/х+3<3/х+2
2) х^3+х^2+х/9х^2-25>=0
3) (х-3)^3(х+1)^3(х+2)^4(3х-2)<0

Answer 1

$mathtt{frac{1}{x+1}+frac{2}{x+3} textless frac{3}{x+2};~frac{(x+2)(x+3)+2(x+1)(x+2)-3(x+1)(x+3)}{(x+1)(x+2)(x+3)} textless 0;}\\mathtt{frac{x^2+5x+6+2x^2+6x+4-3x^2-12x-9}{(x+1)(x+2)(x+3)} textless 0;~frac{1-x}{(x+1)(x+2)(x+3)} textless 0;~frac{x-1}{(x+1)(x+2)(x+3)} textless 0}\\mathtt{+++(-3)---(-2)+++(-1)---(1)+++,~to}\mathtt{to~xin(-3;-2)U(-1;1)}$

$mathtt{frac{x^3+x^2+x}{9x^2-25}geq0;~frac{x(x^2+x+1)}{(3x-5)(3x+5)}geq0;~frac{x}{(3x-5)(3x+5)}geq0}\\mathtt{---(-frac{5}{3})+++[0]---(frac{5}{3})+++,~to~xin(-frac{5}{3};0]U(frac{5}{3};+infty)}$

$mathtt{(x+2)^4(x+1)^3(3x-2)(x-3)^3 textless 0}\\mathtt{---(-2)---(-1)+++(frac{2}{3})---(3)+++,~to}\mathtt{to~xin(-infty;-2)U(-2;-1)U(frac{2}{3};3)}$