Математика: Решить систему уравнений. x-y=6(x+y) x^2-y^2=6.

Решить систему уравнений. x-y=6(x+y) x^2-y^2=6.

Ответы

masha90876 06.10.2020 16:51

(3,5; -2,5),(- 3,5; 2,5)

Пошаговое объяснение:

$\left \{ \begin{array}{lcl} {{x-y=6(x+y),} \\ {x^{2}-y^{2} =6;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{x-y=6x+6y,} \\ {x^{2} -y^{2} =6;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{-5x=7y.} \\ {x^{2} -y^{2} =6;}} \end{array} \right.\Leftrightarrow$

$\Leftrightarrow\left \{ \begin{array}{lcl} {{x=-\frac{7}{5} y,}\\ \\ {(-\frac{7}{5} y)^{2} -y^{2} =6;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{x=-\frac{7}{5} y,} \\ \\{\frac{49}{25}y^{2} -y^{2} =6;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{x=-\frac{7}{5} y,} \\\\ {\frac{24}{25} y^{2} =6;}} \end{array} \right.\Leftrightarrow$

$\Leftrightarrow\left \{ \begin{array}{lcl} {{x=-\frac{7}{5} y,} \\\\ {y^{2} =\frac{25}{4}; }} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{x=-\frac{7}{5}y, } \\\\ \left [ \begin{array}{lcl} {{y=-\frac{5}{2} ,} \\\\ {y=\frac{5}{2}; }} \end{array} \right.{}} \end{array} \right.\Leftrightarrow\left [ \begin{array}{lcl} {{\left \{ \begin{array}{lcl} {{x=3,5,} \\ {y=-2,5};} \end{array} \right.} \\ {\left \{ \begin{array}{lcl} {{x=-3,5,} \\ {y=2,5.}} \end{array} \right.}} \end{array} \right.$

ПОКАЗАТЬ ОТВЕТЫ

Ришат12333323 06.10.2020 16:51

{(-3,5; 2,5); (3,5; -2,5)}

Пошаговое объяснение:

$\displaystyle \left \{ {{x-y=6 \cdot (x+y)} \atop {x^{2} -y^{2} =6}} \right. \Leftrightarrow \left \{ {{x-y=6 \cdot (x+y)} \atop {(x -y) \cdot (x+y) =6}} \right. \Leftrightarrow \\\\\Leftrightarrow \left \{ {{x-y=6 \cdot (x+y)} \atop {6 \cdot (x+y) \cdot (x+y) =6}} \right. \Leftrightarrow \left \{ {{x-y=6 \cdot x+ 6 \cdot y} \atop {(x+y)^{2} =1}} \right. \Leftrightarrow$

$\displaystyle \Leftrightarrow \left \{ {{5 \cdot x = -7 \cdot y} \atop {x+y= б 1}} \right. \Leftrightarrow \left \{ {{x = -\dfrac{7 \cdot y}{5} } \atop {-\dfrac{7 \cdot y}{5} +y= б 1}} \right. \Leftrightarrow \left \{ {{x = -\dfrac{7 \cdot y}{5} } \atop {-\dfrac{2 \cdot y}{5} = б 1}} \right. \Leftrightarrow$

$\displaystyle \Leftrightarrow \left [\begin{array}{cc}\left \{ {{x = -\dfrac{7 \cdot y}{5} } \atop {-\dfrac{2 \cdot y}{5} = -1}} \right.\\\left \{ {{x = -\dfrac{7 \cdot y}{5} } \atop {-\dfrac{2 \cdot y}{5} = 1}} \right.\end{array}\right \Leftrightarrow \left [\begin{array}{ccc}\left \{ {{x = -\dfrac{7}{2} =-3,5} \atop {y=\dfrac{5}{2}=2,5}} \right.\\\left \{ {{x = \dfrac{7}{2} =3,5} \atop {y=-\dfrac{5}{2}=-2,5}} \right.\end{array}\right$