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Question Number 169391 by mathlove last updated on 29/Apr/22

Commented by infinityaction last updated on 29/Apr/22

0

$$\mathrm{0} \\ $$

Commented by mathlove last updated on 29/Apr/22

any solution?

$${any}\:{solution}? \\ $$

Answered by mr W last updated on 30/Apr/22

(x+(√(1+x^2 )))(y+(√(1+y^2 )))=1 (√(1+x^2 ))+x=(√(1+y^2 ))−y x+y=(√(1+y^2 ))−(√(1+x^2 )) xy=1−(√(1+y^2 ))(√(1+x^2 )) 1−xy=(√(1+y^2 ))(√(1+x^2 )) y^2 +x^2 +2xy+(xy)^2 =0 (x+y)^2 +(xy)^2 =0 ⇒xy=0 ⇒x=y=0 ⇒x+y=0 ⇒(x+y)^2 =0

$$\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\left({y}+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\right)=\mathrm{1} \\ $$$$\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+{x}=\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }−{y} \\ $$$${x}+{y}=\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${xy}=\mathrm{1}−\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\mathrm{1}−{xy}=\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${y}^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{2}{xy}+\left({xy}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} +\left({xy}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{xy}=\mathrm{0}\:\Rightarrow{x}={y}=\mathrm{0} \\ $$$$\Rightarrow{x}+{y}=\mathrm{0} \\ $$$$\Rightarrow\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{0} \\ $$

Commented by mathlove last updated on 04/May/22

thank sir

$${thank}\:{sir} \\ $$

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