Question Number 68952 by aliesam last updated on 17/Sep/19
Answered by mind is power last updated on 17/Sep/19
![⇒x−1>0⇒x>1 ⇔((x−1)/(x+1))≤((x+1+3)/(x−1))⇒(x−1)^2 ≤(x+4)(x+1) ⇔x^2 −2x+1≤x^2 +5x+1⇔7x≥0⇒x≥0 S=]1,+∞⌊](https://www.tinkutara.com/question/Q68954.png)
$$\Rightarrow{x}−\mathrm{1}>\mathrm{0}\Rightarrow{x}>\mathrm{1} \\ $$$$\Leftrightarrow\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\leqslant\frac{{x}+\mathrm{1}+\mathrm{3}}{{x}−\mathrm{1}}\Rightarrow\left({x}−\mathrm{1}\right)^{\mathrm{2}} \leqslant\left({x}+\mathrm{4}\right)\left({x}+\mathrm{1}\right) \\ $$$$\Leftrightarrow{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\leqslant{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1}\Leftrightarrow\mathrm{7}{x}\geqslant\mathrm{0}\Rightarrow{x}\geqslant\mathrm{0} \\ $$$$\left.{S}=\right]\mathrm{1},+\infty\lfloor \\ $$