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x-2-2-1-5-y-y-2-2-1-5-x-x-y-x-y-




Question Number 189446 by mathlove last updated on 16/Mar/23
x^2 =(2)^(1/5) +y  y^2 =(2)^(1/5) +x  x∙y=?           x≠y
$${x}^{\mathrm{2}} =\sqrt[{\mathrm{5}}]{\mathrm{2}}+{y} \\ $$$${y}^{\mathrm{2}} =\sqrt[{\mathrm{5}}]{\mathrm{2}}+{x} \\ $$$${x}\centerdot{y}=?\:\:\:\:\:\:\:\:\:\:\:{x}\neq{y} \\ $$
Answered by cortano12 last updated on 16/Mar/23
⇒x^2 −y^2 = y−x  ⇒(x−y)(x+y)+(x−y)=0  ⇒(x−y){x+y+1}=0  ⇒y=−1−x ; x^2 = (2)^(1/5)  −1−x  ⇒x^2 +x=(2)^(1/5)  −1  ∴ xy = x(−1−x)=−x^2 −x=1−(2)^(1/5)
$$\Rightarrow{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\:{y}−{x} \\ $$$$\Rightarrow\left(\mathrm{x}−\mathrm{y}\right)\left(\mathrm{x}+\mathrm{y}\right)+\left(\mathrm{x}−\mathrm{y}\right)=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{x}−\mathrm{y}\right)\left\{\mathrm{x}+\mathrm{y}+\mathrm{1}\right\}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{y}=−\mathrm{1}−\mathrm{x}\:;\:\mathrm{x}^{\mathrm{2}} =\:\sqrt[{\mathrm{5}}]{\mathrm{2}}\:−\mathrm{1}−\mathrm{x} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{x}=\sqrt[{\mathrm{5}}]{\mathrm{2}}\:−\mathrm{1} \\ $$$$\therefore\:\mathrm{xy}\:=\:\mathrm{x}\left(−\mathrm{1}−\mathrm{x}\right)=−\mathrm{x}^{\mathrm{2}} −\mathrm{x}=\mathrm{1}−\sqrt[{\mathrm{5}}]{\mathrm{2}} \\ $$

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