Question Number 151085 by mathdanisur last updated on 18/Aug/21
Answered by talminator2856791 last updated on 18/Aug/21
$$\: \\ $$$$\:\frac{{a}^{\mathrm{2}} −\left({a}+\mathrm{1}\right)^{\mathrm{2}} −\left({a}+\mathrm{2}\right)^{\mathrm{2}} +\left({a}+\mathrm{3}\right)^{\mathrm{2}} }{\left({a}+\mathrm{4}\right)^{\mathrm{2}} −\left({a}+\mathrm{3}\right)^{\mathrm{2}} −\left({a}+\mathrm{2}\right)^{\mathrm{2}} +\left({a}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\: \\ $$$$\:\frac{{a}^{\mathrm{2}} −\left({a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1}\right)−\left({a}^{\mathrm{2}} +\mathrm{4}{a}+\mathrm{4}\right)+\left({a}^{\mathrm{2}} +\mathrm{6}{a}+\mathrm{9}\right)}{\left({a}^{\mathrm{2}} +\mathrm{8}{a}+\mathrm{16}\right)−\left({a}^{\mathrm{2}} +\mathrm{6}{a}+\mathrm{9}\right)−\left({a}^{\mathrm{2}} +\mathrm{4}{a}+\mathrm{4}\right)+\left({a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1}\right)} \\ $$$$\: \\ $$$$\:\frac{\mathrm{4}}{\mathrm{4}}\:=\:\mathrm{1} \\ $$$$\: \\ $$
Commented by mathdanisur last updated on 18/Aug/21
$$\mathrm{Thankyou}\:\mathrm{Ser} \\ $$