Question Number 199259 by hardmath last updated on 30/Oct/23
Commented by Frix last updated on 30/Oct/23
$$\sqrt{\sqrt{\mathrm{12345689654321233}\pm\mathrm{5333334096}\sqrt{\mathrm{12345679}}}}= \\ $$$$=\sqrt{\mathrm{111111127}\pm\mathrm{24}\sqrt{\mathrm{12345679}}}= \\ $$$$=\pm\mathrm{4}+\mathrm{3}\sqrt{\mathrm{12345679}} \\ $$$$\sqrt[{\mathrm{3}}]{{x}−{y}}=\sqrt[{\mathrm{4}}]{\mathrm{8}}=\mathrm{2} \\ $$
Answered by AST last updated on 30/Oct/23
![x^4 +y^4 =2...642466=k;x^4 y^4 =c ⇒(x^2 +y^2 )^2 −2(x^2 y^2 )=k ⇒[(x−y)^2 +2xy]^2 −2(√c)=k⇒(x−y)^2 =(√(k+2(√c)))−2(c)^(1/4) ⇒x−y=(√((√(k+2(√c)))−2(c)^(1/4) )),then take the cube root](https://www.tinkutara.com/question/Q199260.png)
$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} =\mathrm{2}…\mathrm{642466}={k};{x}^{\mathrm{4}} {y}^{\mathrm{4}} ={c} \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} \right)={k} \\ $$$$\Rightarrow\left[\left({x}−{y}\right)^{\mathrm{2}} +\mathrm{2}{xy}\right]^{\mathrm{2}} −\mathrm{2}\sqrt{{c}}={k}\Rightarrow\left({x}−{y}\right)^{\mathrm{2}} =\sqrt{{k}+\mathrm{2}\sqrt{{c}}}−\mathrm{2}\sqrt[{\mathrm{4}}]{{c}} \\ $$$$\Rightarrow{x}−{y}=\sqrt{\sqrt{{k}+\mathrm{2}\sqrt{{c}}}−\mathrm{2}\sqrt[{\mathrm{4}}]{{c}}},{then}\:{take}\:{the}\:{cube}\:{root} \\ $$