Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 164568 by leonhard77 last updated on 19/Jan/22

$$\frac{1}{\sqrt{2-x}}>\frac{1}{x-1}$$

Answered by TheSupreme last updated on 19/Jan/22

$$x<2,x\ne 1$$ $$\{\begin{array}{l}\frac{1}{x-1}<0\cup \{\begin{array}{l}\frac{1}{x-1}>0\\ \frac{1}{2-x}>\frac{1}{{(x-1)}^2}\end{array}\\ \end{array}$$ $$\{\begin{array}{l}x<1\cup \{\begin{array}{l}x>1\\ x^2-x>0\end{array}\\ x<2\end{array}$$ $$x<1\cup 1<x<2$$ $$x<2,x\ne 1$$

Answered by alephzero last updated on 19/Jan/22

$$\frac{1}{\sqrt{2-x}}>\frac{1}{x-1}$$ $$\Rightarrow x\in (-\mathrm{\infty },1)\cup (1,2)..................\left(i\right)$$ $$\frac{1}{\sqrt{2-x}}-\frac{1}{x-1}>0$$ $$\frac{x-1-\sqrt{2-x}}{(x-1)\sqrt{2-x}}>0$$ $$\left(ii\right)\{\begin{array}{l}x-1-\sqrt{2-x}>0\\ (x-1)\sqrt{2-x}>0\end{array}$$ $$\left(iii\right)\{\begin{array}{l}x-1-\sqrt{2-x}<0\\ (x-1)\sqrt{2-x}<0\end{array}$$ $$\left(ii\right)\{\begin{array}{l}x\in (\frac{1+\sqrt{5}}{2},+\mathrm{\infty })\\ x\in (1,2)\cup (2,+\mathrm{\infty })\end{array}$$ $$\left(iii\right)\{\begin{array}{l}x\in (-\mathrm{\infty },\frac{1+\sqrt{5}}{2})\\ x\in (-\mathrm{\infty },1)\end{array}$$ $$\Rightarrow x\in (\frac{1+\sqrt{5}}{2},2)\cup (2,+\mathrm{\infty })...........\left(ii\right)$$ $$x\in (-\mathrm{\infty },1)...........................\left(iii\right)$$ $$\left(i\right)\cup \left(ii\right)\cup \left(iii\right)\Rightarrow$$ $$\Rightarrow x\in (-\mathrm{\infty },1)\cup (\frac{1+\sqrt{5}}{2},2)$$ $$\mathrm{Or}x<1,\frac{1+\sqrt{5}}{2}<x<2$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com