Question

Помогите пожалуйста, то что красной ручкой

Answer 1

1) $\left \{ {{x^{2} + y^{2} = 37} \atop {xy=6}} \right. ; \left \{ {{x^{2} + y^{2} = 37} \atop {x= \frac{6}{y}}} \right. ; \left \{ {{ (\frac{6}{y}) ^{2} + y^{2} =37 } \atop {x=\frac{6}{y}}} \right.$
$(\frac{6}{y}) ^{2} + y^{2} =37$
$\frac{36}{ y^{2} } + y^{2} =37$
ОДЗ: $y^{2} \neq 0$; $y \neq 0$
$36 + y^{4} = 37y^{2}$
$y^{4} - 37y^{2} +36 =0$
Пусть $y^{2} = t, t \geq 0$
$t^{2} - 37t+36=0$
$D = (-37)^{2} -4$×$1$×$36=1369-144=1225= 35^{2} \ \textgreater \ 0 =\ \textgreater \ 2$ корня
$t_{1} = \frac{37+35}{2} = 36$
$t_{2} = \frac{37-35}{2} = 1$
$y^{2} =36 y_{1} = -6; y_{2} = 6$
$y^{2} =1 y_{3} = -1; y_{4} =1$
$x_{1} = \frac{6}{-6} = -1$
$x_{2} = \frac{6}{6} =1$
$x_{3} = \frac{6}{-1} = -6$
$x_{4} = \frac{6}{1} = 6$
Ответ: $y_{1} = -6, x_{1} = -1; y_{2} = -1, x_{2} = -6; y_{3} = 1, x_{3} = 6; y_{4} = 6, x_{4} = 1.$