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Question Number 18798 by chernoaguero@gmail.com last updated on 29/Jul/17

Answered by behi.8.3.4.1.7@gmail.com last updated on 30/Jul/17

$$\frac{x+a}{x+b}=t,\frac{x-a}{x-b}=s\Rightarrow \frac{x^2-a^2}{x^2-b^2}=ts,\frac{a^2+b^2}{ab}=m$$$$\Rightarrow t^2+s^2-mts=0\Rightarrow t=\frac{ms\pm \sqrt{m^2{s}^2-4s^2}}{2}=$$$$\Rightarrow t=\frac{s}{2}(m\pm \sqrt{m^2-4})$$$$m^2-4=\frac{{(a^2+b^2)}^2}{a^2{b}^2}-4=\frac{{(a^2+b^2)}^2-4a^2{b}^2}{a^2{b}^2}=$$$$=\frac{{(a^2-b^2)}^2}{a^2{b}^2}$$$$\Rightarrow m\pm \sqrt{m^2-4}=\frac{a^2+b^2}{ab}\pm \frac{a^2-b^2}{ab}=2\frac{a}{b},2\frac{b}{a}$$$$\Rightarrow \frac{t}{s}=\frac{a}{b},\frac{b}{a}$$$$1)\frac{t}{s}=\frac{a}{b}\Rightarrow \frac{\frac{x+a}{x+b}}{\frac{x-a}{x-b}}=\frac{a}{b}\Rightarrow$$$$\Rightarrow b(x^2+(a-b)x-ab)=a(x^2-(a-b)x-ab)\Rightarrow$$$$(b-a)x^2+(a+b)(a-b)x-ab(a-b)=0$$$$if:b\ne a\Rightarrow x^2-(a+b)x+ab=0\Rightarrow$$$$x=\frac{(a+b)\pm \sqrt{{(a+b)}^2-4ab}}{2}=\frac{(a+b)\pm (a-b)}{2}=a,b$$$$2)\frac{t}{s}=\frac{b}{a}\Rightarrow \frac{\frac{x+a}{x+b}}{\frac{x-a}{x-b}}=\frac{b}{a}\Rightarrow$$$$\Rightarrow a(x^2+(a-b)x-ab)=b(x^2-(a-b)x-ab)\Rightarrow$$$$(a-b)x^2+(a-b)(a+b)x-ab(a-b)=0$$$$\Rightarrow if:a\ne b\Rightarrow x^2+(a+b)x-ab=0$$$$x=\frac{-(a+b)\pm \sqrt{{(a+b)}^2+4ab}}{2}.$$$$if:a=b\Rightarrow t=s\Rightarrow \frac{x+a}{x+a}=\frac{x-a}{x-a}.true:\mathrm{\forall }x.$$

Commented by chernoaguero@gmail.com last updated on 30/Jul/17

$$thankssirireallyappreciateit$$

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