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Question Number 54517 by hassentimol last updated on 05/Feb/19

Prove that (((√(h+1))−1)/h) = (1/((√(h+1))+1)) please...

$$\mathrm{Prove}\:\mathrm{that} \\ $$$$ \\ $$$$\frac{\sqrt{{h}+\mathrm{1}}−\mathrm{1}}{{h}}\:=\:\frac{\mathrm{1}}{\sqrt{{h}+\mathrm{1}}+\mathrm{1}} \\ $$$$ \\ $$$$\mathrm{please}... \\ $$

Answered by Kunal12588 last updated on 05/Feb/19

let (√(h+1))=k h+1=k^2 h=k^2 −1 (((√(h+1))−1)/h)=((k−1)/(k^2 −1))=(((k−1))/((k−1)(k+1)))=(1/(k+1))=(1/((√(h+1))+1))

$${let}\:\sqrt{{h}+\mathrm{1}}={k} \\ $$$${h}+\mathrm{1}={k}^{\mathrm{2}} \\ $$$${h}={k}^{\mathrm{2}} −\mathrm{1} \\ $$$$\frac{\sqrt{{h}+\mathrm{1}}−\mathrm{1}}{{h}}=\frac{{k}−\mathrm{1}}{{k}^{\mathrm{2}} −\mathrm{1}}=\frac{\left({k}−\mathrm{1}\right)}{\left({k}−\mathrm{1}\right)\left({k}+\mathrm{1}\right)}=\frac{\mathrm{1}}{{k}+\mathrm{1}}=\frac{\mathrm{1}}{\sqrt{{h}+\mathrm{1}}+\mathrm{1}} \\ $$

Commented by hassentimol last updated on 05/Feb/19

Thank you Sir God bless you

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$

Answered by math1967 last updated on 05/Feb/19

((((√(h+1))−1)((√(h+1))+1))/(h((√(h+1))+1)))=((h+1−1)/(h((√(h+1))+1))) =(1/((√(h+1))+1))

$$\frac{\left(\sqrt{{h}+\mathrm{1}}−\mathrm{1}\right)\left(\sqrt{{h}+\mathrm{1}}+\mathrm{1}\right)}{{h}\left(\sqrt{{h}+\mathrm{1}}+\mathrm{1}\right)}=\frac{{h}+\mathrm{1}−\mathrm{1}}{{h}\left(\sqrt{{h}+\mathrm{1}}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\sqrt{{h}+\mathrm{1}}+\mathrm{1}} \\ $$

Commented by hassentimol last updated on 05/Feb/19

Thank you Sir

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$

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