Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 121446 by mr W last updated on 08/Nov/20

Answered by liberty last updated on 08/Nov/20

u = x+(√(1+x^2 )) u(y+(√(1+y^2 )) ) = 1 ⇒(√(1+y^2 )) = (1/u)−y ⇒1+y^2 =(1/u^2 )−2(y/u)+y^2 ((2y)/u) = (1/u^2 )−u ⇒y=(1/2)((1/u)−u) (1/u)=(1/(x+(√(1+x^2 )))) = −x+(√(1+x^2 )) y=(1/2)(−x+(√(1+x^2 ))−x−(√(1+x^2 )) ) y=−x⇒x+y =0 ∧(x+y)^2 =0

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

Answered by MJS_new last updated on 08/Nov/20

0 let x=((p^2 −1)/(2p))∧y=((q^2 −1)/(2q)) ⇒ pq=1 ⇒ y=−((p^2 −1)/(2p)) ⇒ x+y=0

$$\mathrm{0} \\ $$$$\mathrm{let}\:{x}=\frac{{p}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{p}}\wedge{y}=\frac{{q}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{q}} \\ $$$$\Rightarrow\:{pq}=\mathrm{1}\:\Rightarrow\:{y}=−\frac{{p}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{p}}\:\Rightarrow\:{x}+{y}=\mathrm{0} \\ $$

Answered by mr W last updated on 08/Nov/20

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com