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Question-200265




Question Number 200265 by cherokeesay last updated on 16/Nov/23
Answered by witcher3 last updated on 16/Nov/23
 { (((x^2 −y)^2 +5^2 −2.5(x^2 −y)=0)),(((2)⇔(2))) :}  ⇔ { (((x^2 −y−5)^2 =0)),(((√((x^2 −3)^2 +4))−(√(2x^2 +y^2 −6))=0)) :}  ⇔ { ((y=x^2 −5)),(((x^4 −6x^2 +13)=2x^2 +y^2 −6)) :}  x^4 −8x^2 +19−(x^4 −10x^2 +25)=0  2x^2 −6=0  x=(√3)  x=−(√3)  y=−2  S={((√3),−2);(−(√3),−2)}
$$\begin{cases}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}\right)^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{2}.\mathrm{5}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}\right)=\mathrm{0}}\\{\left(\mathrm{2}\right)\Leftrightarrow\left(\mathrm{2}\right)}\end{cases} \\ $$$$\Leftrightarrow\begin{cases}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{0}}\\{\sqrt{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{3}\right)^{\mathrm{2}} +\mathrm{4}}−\sqrt{\mathrm{2x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{6}}=\mathrm{0}}\end{cases} \\ $$$$\Leftrightarrow\begin{cases}{\mathrm{y}=\mathrm{x}^{\mathrm{2}} −\mathrm{5}}\\{\left(\mathrm{x}^{\mathrm{4}} −\mathrm{6x}^{\mathrm{2}} +\mathrm{13}\right)=\mathrm{2x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{6}}\end{cases} \\ $$$$\mathrm{x}^{\mathrm{4}} −\mathrm{8x}^{\mathrm{2}} +\mathrm{19}−\left(\mathrm{x}^{\mathrm{4}} −\mathrm{10x}^{\mathrm{2}} +\mathrm{25}\right)=\mathrm{0} \\ $$$$\mathrm{2x}^{\mathrm{2}} −\mathrm{6}=\mathrm{0} \\ $$$$\mathrm{x}=\sqrt{\mathrm{3}} \\ $$$$\mathrm{x}=−\sqrt{\mathrm{3}} \\ $$$$\mathrm{y}=−\mathrm{2} \\ $$$$\mathrm{S}=\left\{\left(\sqrt{\mathrm{3}},−\mathrm{2}\right);\left(−\sqrt{\mathrm{3}},−\mathrm{2}\right)\right\} \\ $$$$ \\ $$

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