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Question Number 45327 by Necxx last updated on 12/Oct/18

Answered by rahul 19 last updated on 12/Oct/18

$$x=1+\frac{1}{3+\frac{1}{1+x}}$$$$\Rightarrow x=1+\frac{1+x}{4+3x}$$$$\Rightarrow 4x+3x^2=5+4x$$$$\Rightarrow x=\underset{-}{+}\sqrt{\frac{5}{3}.}\Rightarrow \left(d\right).$$

Commented by Necxx last updated on 12/Oct/18

$$thanksboss$$

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

$$\frac{1}{x-1}=3+\frac{1}{2+\frac{1}{3+\frac{1}{2+\frac{1}{3+\frac{1}{2...\mathrm{\infty }}}}}}$$$$k=3+\frac{1}{2+\frac{1}{k}}letk=\frac{1}{x-1}$$$$k=3+\frac{k}{2k+1}$$$$k(2k+1)=6k+3+k$$$$2k^2+k-7k-3=0$$$$2k^2-6k-3=0$$$$\frac{2}{{(x-1)}^2}-\frac{6}{x-1}-3=0$$$$2-6(x-1)-3(x^2-2x+1)=0$$$$-3x^2+6x-3+2-6x+6=0$$$$-3x^2=-5$$$$x=\pm \sqrt{\frac{5}{3}}$$

Commented by Necxx last updated on 12/Oct/18

$$thanksboss$$

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