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Question Number 180785 by Vynho last updated on 17/Nov/22

solve (1−x−x^2 ...)(2−x−x^2 ...)

solve(1−x−x2...)(2−x−x2...)

Answered by Rasheed.Sindhi last updated on 17/Nov/22

( 1−(x+x^2 +...) )( 2−(x+x^2 +...) ) If ∣x∣<1: (1−(x/(1−x)))(2−(x/(1−x))) ((1−x−x)/(1−x))∙((2−2x−x)/(1−x)) =(((1−2x)(2−3x))/((1−x)^2 ))=((6x^2 −7x+2)/(x^2 −2x+1))

(1−(x+x2+...))(2−(x+x2+...))If∣x∣<1:(1−x1−x)(2−x1−x)1−x−x1−x⋅2−2x−x1−x=(1−2x)(2−3x)(1−x)2=6x2−7x+2x2−2x+1

Answered by mr W last updated on 17/Nov/22

let t=x+x^2 +...=x(1+x+x^2 +...) ⇒(1−t)(2−t)=0 ⇒t=1 or 2 ⇒x(1+x+x^2 +...)=1 or 2 x<1 (x/(1−x))=1 or 2 ⇒x=(1/2) or (2/3)

lett=x+x2+...=x(1+x+x2+...)⇒(1−t)(2−t)=0⇒t=1or2⇒x(1+x+x2+...)=1or2x<1x1−x=1or2⇒x=12or23

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