X'2+y'2=50 xy=7 решить систему

Question 1

Question 2

Ответ:

(1;7), (7;1) , (-1;-7) , (-7;-1).

Объяснение:

$\left \{ \begin{array}{lcl} {{x^{2}+y^{2}=50, } \\ {xy=7;}} \end{array} \right. \Leftrightarrow\left \{ \begin{array}{lcl} {{x^{2}+y^{2} =50, } \\ {2xy=14;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{x^{2}+2xy+y^{2} =64, } \\ {xy=7;}} \end{array} \right. \Leftrightarrow$

$\left \{ \begin{array}{lcl} {{(x+y)^{2} =64,} \\ {xy=7}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{\left [ \begin{array}{lcl} {{x+y=8,} \\ {x+y=-8,}} \end{array} \right.} \\ {xy=7.}} \end{array} \right.$

Решим первую систему:

$\left \{ \begin{array}{lcl} {{x+y=8,} \\ {xy=7;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{y=8-x,} \\ {x(8-x) =7;}} \end{array} \right.\Leftrightarrow$

$\left \{ \begin{array}{lcl} {{y=8-x,} \\ {x^{2} -8x+7=0;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{y=8-x} \\ {\left [\begin{array}{lcl} {{x=1} \\ {x=7}} \end{array} \right.}} \end{array}$ $\Leftrightarrow\left [ \begin{array}{lcl} {\left \{ \begin{array}{lcl} {{x=1,} \\ {y=7;}} \end{array} \right{} \\ {\left \{ \begin{array}{lcl} {{x=7,} \\ {y=1;}} \end{array} \right}} \end{array} \right$

Решим вторую систему:

$\left \{ \begin{array}{lcl} {{x+y=-8,} \\ {xy=7;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{y=-8-x} \\ {x(-8-x)=7;}} \end{array} \right.\Leftrightarrow \left \{ \begin{array}{lcl} {{y=-8-x,} \\ {x^{2} +x+7=0;}} \end{array} \right. \Leftrightarrow$

$\left \{ \begin{array}{lcl} {{y=-8-x,} \\ {\left [ \begin{array}{lcl} {{x=-1,} \\ {x=-7;}} \end{array} \right.}} \end{array} \right. \Leftrightarrow \left [ \begin{array}{lcl} {{\left \{ \begin{array}{lcl} {{x=-1,} \\ {y=-7;}} \end{array} \right.} \\ {\left \{ \begin{array}{lcl} {{x=-7,} \\ {y=-1.}} \end{array} \right.}} \end{array} \right.$