Question

Resolva as inequações $\frac{x - 1}{x -2} \ \textless \ \frac{x - 3}{x - 4}$
$\frac{x}{x -1} - \frac{2}{x + 1}$ ≤0

Answer 1

$a)\ \dfrac{x-1}{x-2}\ \textless \ \dfrac{x-3}{x-4}\\\\\\\dfrac{(x-4)(x-1)-(x-2)(x-3)}{(x-2)(x-4)}\ \textless \ 0\\\\\\\dfrac{(x^2-5x+4)-(x^2-5x+6)}{(x-2)(x-4)}\ \textless \ 0\\\\\\\dfrac{x^2-5x+4-x^2+5x-6}{(x-2)(x-4)}\ \textless \ 0\\\\\\\dfrac{-2}{(x-2)(x-4)}\ \textless \ 0\\\\\\-\dfrac{2}{(x-2)(x-4)}\ \textless \ 0\\\\\\\dfrac{2}{(x-2)(x-4)}\ \textgreater \ 0\\\\\\\therefore\ (x-2)(x-4)\ \textgreater \ 0\\\\\\\\\begin{matrix}\alpha)&x-2\ \textgreater \ 0&\\\\&x\ \textgreater \ 2&\text{ou}\ \ ]2,+\infty[\end{matrix}\qquad\qquad\qquad\begin{matrix}\beta)&x-4\ \textgreater \ 0&\\\\&x\ \textgreater \ 4&\text{ou}\ \ ]4,+\infty[\end{matrix}$


$\alpha)\times\beta):\ \ ]2,+\infty[\ \ \times\ \ ]4,+\infty[\ \ =\ \ ]\!\!-\infty,2[\ \ \cup\ \ ]4,+\infty[\\\\\text{ou}:\ \{x\in\mathbb{R}\mid{x}\ \textless \ 2\text{ ou }x\ \textgreater \ 4\}\\\\\\\boxed{S=\{x\in\mathbb{R}\mid{x}\ \textless \ 2\text{ ou }x\ \textgreater \ 4\}}$




$b)\ \dfrac{x}{x-1}-\dfrac{2}{x+1}\le0\\\\\\\dfrac{(x+1)x-(x-1)2}{(x-1)(x+1)}\le0\\\\\\\dfrac{(x^2+x)-(2x-2)}{(x-1)(x+1)}\le0\\\\\\\dfrac{x^2+x-2x+2}{(x-1)(x+1)}\le0\\\\\\\dfrac{x^2-x+2}{(x-1)(x+1)}\le0\\\\\\\\x^2-x+2\le0\\\\\Delta=-7\\\\\therefore\ \nexists{x}\mid{x}^2-x+2\le0\\\\\\\begin{matrix}\alpha)&x-1\ \textless \ 0&\\\\&x\ \textless \ 1&\text{ou}\ \ ]\!\!-\infty,1[\end{matrix}\qquad\qquad\begin{matrix}\beta)&x+1\ \textless \ 0&\\\\&x\ \textless \ -1&\text{ou}\ \ ]\!\!-\infty,-1[\end{matrix}$



$\alpha)\times\beta):\ \ ]\!\!-\infty,1[\ \ \times\ \ ]\!\!-\infty,-1[\ \ =\ \ ]\!\!-1,1[\\\\\text{ou: }\{x\in\mathbb{R}\mid-1\ \textless \ x\ \textless \ 1\}\\\\\\\boxed{S=\{x\in\mathbb{R}\mid-1\ \textless \ x\ \textless \ 1\}}$