Question

Решите способом введения новой переменной систему уравнений:

Answer 1

$\left\{\begin{array}{ccc}(u+v)^2-5(u+v)+4=0\\(u-v)^2-(u-v)-2=0\end{array}\right\ \ t=u+v\ ,\ \ p=u-v\ \left\{\begin{array}{ccc}t^2-5t+4=0\\p^2-p-2=0\end{array}\right\\\\\\\left\{\begin{array}{l}t_1=1\ ,\ t_2=4\\p_1=-1\ ,\ p_2=2\end{array}\right\ \ \left\{\begin{array}{ccc}u+v=1\ \ ,\ \ u+v=4\\u-v=-1\ ,\ u-v=2\end{array}\right$

$1)\ \ \left\{\begin{array}{l}u+v=1\\u-v=-1\end{array}\right\ \ \left\{\begin{array}{l}2u=0\\2v=2\end{array}\right\ \ \left\{\begin{array}{l}u=0\\v=1\end{array}\right\\\\\\2)\ \ \left\{\begin{array}{l}u+v=1\\u-v=2\end{array}\right\ \ \left\{\begin{array}{l}2u=3\\2v=-1\end{array}\right\ \ \left\{\begin{array}{l}u=1,5\\v=-0,5\end{array}\right$

$3)\ \ \left\{\begin{array}{l}u+v=4\\u-v=-1\end{array}\right\ \ \left\{\begin{array}{l}2u=3\\2v=5\end{array}\right\ \ \left\{\begin{array}{l}u=1,5\\v=2,5\end{array}\right\\\\\\4)\ \ \left\{\begin{array}{l}u+v=4\\u-v=2\end{array}\right\ \ \left\{\begin{array}{l}2u=6\\2v=2\end{array}\right\ \ \left\{\begin{array}{l}u=3\\v=1\end{array}\right\\\\\\Otvet:\ \ (\, 0\, ;\, 1\, )\ ,\ (\, 1,5\, ;\, -0,5\, )\ ,\ (\, 1,5\, ;\, 2,5\, )\ ,\ (\, 3\, ;\, 1\, )\ .$

Answer 2

Ответ:

Объяснение:

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