{x кв+y кв=25 {x+y=7

[latex] \left \{ {{x^{2}+y^{2}=25} \atop {x+y=7}} \right. [/latex] [latex] \left \{ {{x^{2}+y^{2}=25} \atop {(x+y)^{2}=7^{2}}} \right. [/latex] [latex] \left \{ {{x^{2}+y^{2}=25} \atop {x^{2}+y^{2}+2yx=49}} \right. [/latex] [latex] \left \{ {{x^{2}+y^{2}=25} \atop {25+2yx=49}} \right. [/latex] [latex] \left \{ {{x^{2}+y^{2}=25} \atop {2yx=24}} \right. [/latex] [latex] \left \{ {{x^{2}+y^{2}=25} \atop {yx=12}} \right. [/latex] [latex] \left \{ {{x^{2}+y^{2}=25} \atop {x= \frac{12}{y}}} \right. [/latex] [latex] \left \{ {{(\frac{12}{y})^{2}+y^{2}=25} \atop {x= \frac{12}{y}}} \right. [/latex] x>0, y>0 [latex]\frac{144}{y^{2}}+y^{2}=25[/latex] [latex]\frac{144+y^{4}}{y^{2}}=25[/latex] [latex]144+y^{4}=25y^{2}[/latex] [latex]y^{4}-25y^{2}+144=0[/latex] Замена: [latex]y^{2}=t\ \textgreater \ 0[/latex] [latex]t^{2}-25t+144=0, D=25^{2}-4*144=49[/latex] [latex]t_{1}= \frac{25-7}{2}=9\ \textgreater \ 0[/latex] [latex]t_{2}= \frac{25+7}{2}=16\ \textgreater \ 0[/latex] Вернемся к замене: 1) [latex]y^{2}=9[/latex] [latex]y_{1}=3[/latex] [latex]x_{1}= \frac{12}{y_{1}}=4[/latex] [latex]y_{2}=-3[/latex] - посторонний корень 2) [latex]y^{2}=16[/latex] [latex]y_{3}=4[/latex] [latex]x_{3}= \frac{12}{y_{1}}=3[/latex] [latex]y_{4}=-4[/latex] - посторонний корень Ответ: (3;4), (4;3)