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Question Number 188079 by 073 last updated on 25/Feb/23

Commented by 073 last updated on 25/Feb/23

f((x/(x^2 +x+1)))=(x^2 /(x^4 +x^2 +x)) f(x)=? please solution

$$\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\right)=\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=? \\ $$$$\mathrm{please}\:\mathrm{solution} \\ $$

Commented by mnjuly1970 last updated on 25/Feb/23

(x/(x^( 2) + x +1)) =t ⇒ ((x^( 2) +x +1)/x) =(1/t) x + (1/x) = ((1−t)/t) ⇒^∧^( 2) x^( 2) + (1/(x^( 2) )) +2=((1−2t+t^( 2) )/t^( 2) ) x^( 2) + (1/x^( 2 ) ) +1= ((1−2t)/t^( 2) ) ((x^( 4) + x^( 2) +1)/x^( 2) ) = ((1−2t)/t^( 2) ) (x^( 2) /(1+ x^( 2) +x^( 4) )) = (t^( 2) /(1−2t)) ∴ f (t )= (t^( 2) /(1−2t))

$$\:\:\:\:\frac{{x}}{{x}^{\:\mathrm{2}} \:+\:{x}\:+\mathrm{1}}\:={t}\:\Rightarrow\:\frac{{x}^{\:\mathrm{2}} +{x}\:+\mathrm{1}}{{x}}\:=\frac{\mathrm{1}}{{t}} \\ $$$$\:\:\:\:\:\:{x}\:+\:\frac{\mathrm{1}}{{x}}\:=\:\frac{\mathrm{1}−{t}}{{t}}\:\overset{\wedge^{\:\mathrm{2}} } {\Rightarrow}\:{x}^{\:\mathrm{2}} \:+\:\frac{\mathrm{1}}{{x}^{\:\mathrm{2}} \:}\:+\mathrm{2}=\frac{\mathrm{1}−\mathrm{2}{t}+{t}^{\:\mathrm{2}} }{{t}^{\:\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:{x}^{\:\mathrm{2}} \:+\:\frac{\mathrm{1}}{{x}^{\:\mathrm{2}\:} }\:+\mathrm{1}=\:\frac{\mathrm{1}−\mathrm{2}{t}}{{t}^{\:\mathrm{2}} }\: \\ $$$$\:\:\:\:\:\:\frac{{x}^{\:\mathrm{4}} \:+\:{x}^{\:\mathrm{2}} \:+\mathrm{1}}{{x}^{\:\mathrm{2}} }\:=\:\frac{\mathrm{1}−\mathrm{2}{t}}{{t}^{\:\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\frac{{x}^{\:\mathrm{2}} }{\mathrm{1}+\:{x}^{\:\mathrm{2}} \:+{x}^{\:\mathrm{4}} }\:=\:\frac{{t}^{\:\mathrm{2}} }{\mathrm{1}−\mathrm{2}{t}} \\ $$$$\:\:\:\:\therefore\:\:\:{f}\:\left({t}\:\right)=\:\frac{{t}^{\:\mathrm{2}} }{\mathrm{1}−\mathrm{2}{t}} \\ $$

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