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

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

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

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