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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

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

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