Question-120166

Question Number 120166 by Algoritm last updated on 29/Oct/20

Commented by Algoritm last updated on 29/Oct/20

$$\left.\mathrm{what}\:?\:\right): \\ $$

Answered by nimnim last updated on 30/Oct/20

a+b+c+d=S..................(1) (1/a)+(1/b)+(1/c)+(1/d)=S..........(2) (1/(a−1))+(1/(b−1))+(1/(c−1))+(1/(d−1))=S......(3) a+(1/a)+(1/(a−1))=((a(a−1)+a−1+1)/(a(a−1)))=(a^2 /(a(a−1)))=(a/(a−1)) ∴ (a/(a−1))+(b/(b−1))+(c/(c−1))+(d/(d−1))=3S ⇒((a/(a−1))−1)+((b/(b−1))−1)+((c/(c−1))−1)+((d/(d−1))−1)=3S−4 ⇒(1/(a−1))+(1/(b−1))+(1/(c−1))+(1/(d−1))=3S−4......(4) From (3) and (4) 3S−4=S ⇒S=2★

$$\:\:{a}+{b}+{c}+{d}={S}………………\left(\mathrm{1}\right) \\ $$$$\:\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\frac{\mathrm{1}}{{d}}={S}……….\left(\mathrm{2}\right) \\ $$$$\:\frac{\mathrm{1}}{{a}−\mathrm{1}}+\frac{\mathrm{1}}{{b}−\mathrm{1}}+\frac{\mathrm{1}}{{c}−\mathrm{1}}+\frac{\mathrm{1}}{{d}−\mathrm{1}}={S}……\left(\mathrm{3}\right) \\ $$$$ \\ $$$${a}+\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{a}−\mathrm{1}}=\frac{{a}\left({a}−\mathrm{1}\right)+{a}−\mathrm{1}+\mathrm{1}}{{a}\left({a}−\mathrm{1}\right)}=\frac{{a}^{\mathrm{2}} }{{a}\left({a}−\mathrm{1}\right)}=\frac{{a}}{{a}−\mathrm{1}} \\ $$$$\therefore\:\frac{{a}}{{a}−\mathrm{1}}+\frac{{b}}{{b}−\mathrm{1}}+\frac{{c}}{{c}−\mathrm{1}}+\frac{{d}}{{d}−\mathrm{1}}=\mathrm{3}{S} \\ $$$$\Rightarrow\left(\frac{{a}}{{a}−\mathrm{1}}−\mathrm{1}\right)+\left(\frac{{b}}{{b}−\mathrm{1}}−\mathrm{1}\right)+\left(\frac{{c}}{{c}−\mathrm{1}}−\mathrm{1}\right)+\left(\frac{{d}}{{d}−\mathrm{1}}−\mathrm{1}\right)=\mathrm{3}{S}−\mathrm{4} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}−\mathrm{1}}+\frac{\mathrm{1}}{{b}−\mathrm{1}}+\frac{\mathrm{1}}{{c}−\mathrm{1}}+\frac{\mathrm{1}}{{d}−\mathrm{1}}=\mathrm{3}{S}−\mathrm{4}……\left(\mathrm{4}\right) \\ $$$${From}\:\left(\mathrm{3}\right)\:{and}\:\left(\mathrm{4}\right) \\ $$$$\mathrm{3}{S}−\mathrm{4}={S} \\ $$$$\Rightarrow{S}=\mathrm{2}\bigstar \\ $$$$ \\ $$

Commented by Algoritm last updated on 30/Oct/20