Question Number 183019 by mathlove last updated on 18/Dec/22
$${Q}\:\:\:{f}\left({x}\right)={a}^{{x}} +{b}^{{x}} \:\:\:\:\:{g}\left({x}\right)=\frac{{f}\left({x}\right)}{{f}\left({x}−\mathrm{2}\right)} \\ $$$${g}\left(\mathrm{3}\right)=? \\ $$$$ \\ $$
Commented by mathlove last updated on 18/Dec/22
$${cheke}\:{this}?? \\ $$
Commented by lahcene1969 last updated on 18/Dec/22
Answered by mathlove last updated on 18/Dec/22
Commented by HeferH last updated on 18/Dec/22
$$\:\frac{{a}^{{x}} \:+\:{b}^{{x}} }{\frac{{a}^{{x}} {b}^{\mathrm{2}} +{b}^{{x}} {a}^{\mathrm{2}} }{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }}\:=\:\frac{\left({a}^{{x}} +{b}^{{x}} \right)\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} \right)}{{a}^{{x}} {b}^{\mathrm{2}} \:+\:{b}^{{x}} {a}^{\mathrm{2}} }\: \\ $$$$\: \\ $$
Commented by JDamian last updated on 18/Dec/22
brackets are needed
Answered by qaz last updated on 18/Dec/22
$${g}\left(\mathrm{3}\right)=\frac{{f}\left(\mathrm{3}\right)}{{f}\left(\mathrm{1}\right)}=\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} }{{a}+{b}}={a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \\ $$
Commented by mathlove last updated on 18/Dec/22
$${whear}\:{place}\:{is}\:{wrong}\:{my}\: \\ $$$${solution}?? \\ $$
Commented by JDamian last updated on 18/Dec/22
$${g}\left({x}\right)\:=\:\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({a}^{{x}} +{b}^{{x}} \right)}{{b}^{\mathrm{2}} {a}^{{x}} +{a}^{\mathrm{2}} {b}^{{x}} } \\ $$
Answered by lahcene1969 last updated on 18/Dec/22
