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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+(1/(3+(1/(1+x)))) ⇒x=1+((1+x)/(4+3x)) ⇒4x+3x^2 =5+4x ⇒x= +_− (√((5/3).)) ⇒ (d).

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

Commented by Necxx last updated on 12/Oct/18

thanks boss

$${thanks}\:{boss} \\ $$

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

(1/(x−1))=3+(1/(2+(1/(3+(1/(2+(1/(3+(1/(2...∞)))))))))) k=3+(1/(2+(1/k))) let k=(1/(x−1)) k=3+(k/(2k+1)) k(2k+1)=6k+3+k 2k^2 +k−7k−3=0 2k^2 −6k−3=0 (2/((x−1)^2 ))−(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=±(√(5/3))

$$\frac{\mathrm{1}}{{x}−\mathrm{1}}=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{3}+\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{3}+\frac{\mathrm{1}}{\mathrm{2}...\infty}}}}} \\ $$$${k}=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{1}}{{k}}}\:\:\:\:\:\:\:{let}\:{k}=\frac{\mathrm{1}}{{x}−\mathrm{1}} \\ $$$${k}=\mathrm{3}+\frac{{k}}{\mathrm{2}{k}+\mathrm{1}} \\ $$$${k}\left(\mathrm{2}{k}+\mathrm{1}\right)=\mathrm{6}{k}+\mathrm{3}+{k} \\ $$$$\mathrm{2}{k}^{\mathrm{2}} +{k}−\mathrm{7}{k}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{2}{k}^{\mathrm{2}} −\mathrm{6}{k}−\mathrm{3}=\mathrm{0} \\ $$$$\frac{\mathrm{2}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{6}}{{x}−\mathrm{1}}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{2}−\mathrm{6}\left({x}−\mathrm{1}\right)−\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$−\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{3}+\mathrm{2}−\mathrm{6}{x}+\mathrm{6}=\mathrm{0} \\ $$$$−\mathrm{3}{x}^{\mathrm{2}} =−\mathrm{5} \\ $$$${x}=\pm\sqrt{\frac{\mathrm{5}}{\mathrm{3}}} \\ $$

Commented by Necxx last updated on 12/Oct/18

thanks boss

$${thanks}\:{boss} \\ $$

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