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x-a-2-1-1-x-b-2-1-2-1-x-a-b-1-x-b-a-1-2-x-a-x-b-2-1-2-1-x-a-b-2-1-2-1-1-x-a-b-2-1-2-1-




Question Number 167492 by mathlove last updated on 18/Mar/22
x^a =(√2)+1     .........(1)   x^b =(√2)−1     .........(2)  (1/x^(a−b) )+(1/x^(b−a) )=?  (1)÷(2)⇒(x^a /x^b )=(((√2)+1)/( (√(2−1))))⇒x^(a−b) =(((√2)+1)/( (√(2−1))))⇒(1/x^(a−b) )=(((√2)−1)/( (√2)+1))....(3)  (2)÷(1)⇒(x^b /x^a )=(((√2)−1)/( (√2)+1))⇒x^(b−a) =(((√2)−1)/( (√2)+1))⇒(1/x^(b−a) )=(((√2)+1)/( (√2)−1))....(4)  (1/x^(a−b) )+(1/x^(b−a) )=(((√2)−1)/( (√2)+1))+(((√2)+1)/( (√2)−1))=((((√2)−1)((√2)−1)+((√2)+1)((√2)+1))/(((√2)+1)((√2)−1)))  =((((√2)−1)^2 +((√2)+1)^2 )/(2−1))=2−2(√2)+1+2+2(√2)+1  (1/x^(a−b) )+(1/x^(b−a) )=6
$${x}^{{a}} =\sqrt{\mathrm{2}}+\mathrm{1}\:\:\:\:\:………\left(\mathrm{1}\right)\: \\ $$$${x}^{{b}} =\sqrt{\mathrm{2}}−\mathrm{1}\:\:\:\:\:………\left(\mathrm{2}\right) \\ $$$$\frac{\mathrm{1}}{{x}^{{a}−{b}} }+\frac{\mathrm{1}}{{x}^{{b}−{a}} }=? \\ $$$$\left(\mathrm{1}\right)\boldsymbol{\div}\left(\mathrm{2}\right)\Rightarrow\frac{{x}^{{a}} }{{x}^{{b}} }=\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{2}−\mathrm{1}}}\Rightarrow{x}^{{a}−{b}} =\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{2}−\mathrm{1}}}\Rightarrow\frac{\mathrm{1}}{{x}^{{a}−{b}} }=\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}….\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{2}\right)\boldsymbol{\div}\left(\mathrm{1}\right)\Rightarrow\frac{{x}^{{b}} }{{x}^{{a}} }=\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}\Rightarrow{x}^{{b}−{a}} =\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}\Rightarrow\frac{\mathrm{1}}{{x}^{{b}−{a}} }=\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}….\left(\mathrm{4}\right) \\ $$$$\frac{\mathrm{1}}{{x}^{{a}−{b}} }+\frac{\mathrm{1}}{{x}^{{b}−{a}} }=\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}+\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}=\frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)+\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}{\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)} \\ $$$$=\frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}−\mathrm{1}}=\mathrm{2}−\cancel{\mathrm{2}\sqrt{\mathrm{2}}}+\mathrm{1}+\mathrm{2}+\cancel{\mathrm{2}\sqrt{\mathrm{2}}}+\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{x}^{{a}−{b}} }+\frac{\mathrm{1}}{{x}^{{b}−{a}} }=\mathrm{6} \\ $$
Answered by Rasheed.Sindhi last updated on 18/Mar/22
x^a =(√2)+1     .........(1)   x^b =(√2)−1     .........(2)  (1/x^(a−b) )+(1/x^(b−a) )=?  (1/x^(a−b) )+(1/x^(b−a) )=(1/(  (x^a /x^b )  ))+(1/(  (x^b /x^a )  ))=(x^b /x^a )+(x^a /x^b )  =(((x^b )^2 +(x^a )^2 )/(x^a x^b ))  =((((√2) −1)^2 +((√2) +1)^2 )/(((√2) +1)((√2) −1)))  =((2((√2) )^2 +2(1)^2 )/(((√2) )^2 −(1)^2 ))  =((4+2)/(2−1))  =6
$${x}^{{a}} =\sqrt{\mathrm{2}}+\mathrm{1}\:\:\:\:\:………\left(\mathrm{1}\right)\: \\ $$$${x}^{{b}} =\sqrt{\mathrm{2}}−\mathrm{1}\:\:\:\:\:………\left(\mathrm{2}\right) \\ $$$$\frac{\mathrm{1}}{{x}^{{a}−{b}} }+\frac{\mathrm{1}}{{x}^{{b}−{a}} }=? \\ $$$$\frac{\mathrm{1}}{{x}^{{a}−{b}} }+\frac{\mathrm{1}}{{x}^{{b}−{a}} }=\frac{\mathrm{1}}{\:\:\frac{{x}^{{a}} }{{x}^{{b}} }\:\:}+\frac{\mathrm{1}}{\:\:\frac{{x}^{{b}} }{{x}^{{a}} }\:\:}=\frac{{x}^{{b}} }{{x}^{{a}} }+\frac{{x}^{{a}} }{{x}^{{b}} } \\ $$$$=\frac{\left({x}^{{b}} \right)^{\mathrm{2}} +\left({x}^{{a}} \right)^{\mathrm{2}} }{{x}^{{a}} {x}^{{b}} } \\ $$$$=\frac{\left(\sqrt{\mathrm{2}}\:−\mathrm{1}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}\:+\mathrm{1}\right)^{\mathrm{2}} }{\left(\sqrt{\mathrm{2}}\:+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}\:−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{2}\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{1}\right)^{\mathrm{2}} }{\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} −\left(\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}+\mathrm{2}}{\mathrm{2}−\mathrm{1}} \\ $$$$=\mathrm{6} \\ $$

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