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Question-107315




Question Number 107315 by Algoritm last updated on 10/Aug/20
Commented by bemath last updated on 10/Aug/20
(1/(2020−(√(2020^2 −1))))= a+b−2(√(ab))  ((2020+(√(2020^2 −1)))/(2020^2 −(2020^2 −1))) = a+b−2(√(ab))  2020+(√(2020^2 −1)) = a+b−2(√(ab))  → { ((a+b=2020)),(((√(2020^2 −1)) =−2(√(ab)))) :}  ?? the question is right?  i think it (1/( (√(2020−(√(2020^2 −1))))))=(√a)+(√b)
$$\frac{\mathrm{1}}{\mathrm{2020}−\sqrt{\mathrm{2020}^{\mathrm{2}} −\mathrm{1}}}=\:{a}+{b}−\mathrm{2}\sqrt{{ab}} \\ $$$$\frac{\mathrm{2020}+\sqrt{\mathrm{2020}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{2020}^{\mathrm{2}} −\left(\mathrm{2020}^{\mathrm{2}} −\mathrm{1}\right)}\:=\:{a}+{b}−\mathrm{2}\sqrt{{ab}} \\ $$$$\mathrm{2020}+\sqrt{\mathrm{2020}^{\mathrm{2}} −\mathrm{1}}\:=\:{a}+{b}−\mathrm{2}\sqrt{{ab}} \\ $$$$\rightarrow\begin{cases}{{a}+{b}=\mathrm{2020}}\\{\sqrt{\mathrm{2020}^{\mathrm{2}} −\mathrm{1}}\:=−\mathrm{2}\sqrt{{ab}}}\end{cases} \\ $$$$??\:{the}\:{question}\:{is}\:{right}? \\ $$$${i}\:{think}\:{it}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2020}−\sqrt{\mathrm{2020}^{\mathrm{2}} −\mathrm{1}}}}=\sqrt{{a}}+\sqrt{{b}} \\ $$

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