Алгебра: с алгеброй, только 2 задание (не предлагайте photomath)

с алгеброй, только 2 задание (не предлагайте photomath)

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Ответ:

GNOM888

15.10.2020 17:08

$1)\ \ \left\{\begin{array}{ccc}x^2+y^2=7\\3x^2-y^2=9\end{array}\right\ \oplus \ \left\{\begin{array}{l}x^2+y^2=7\\4x^2=16\end{array}\right\ \ \left\{\begin{array}{l}y^2=7-x^2\\x^2=4\end{array}\right\ \ \left\{\begin{array}{l}y^2=3\\x=\pm 2\end{array}\right\\\\\\\left\{\begin{array}{l}y=\pm \sqrt3\\x=\pm 2\end{array}\right\ \ \ Otvet:\ \ (-2,-\sqrt3)\ ,\ (-2,\sqrt3)\ ,\ (2,-\sqrt3)\ ,\ (2,\sqrt3)\ .$

$2)\ \ \left\{\begin{array}{ccc}2x^2-1=y^2\\3y^2=2x^2-1\end{array}\right\ \ \left\{\begin{array}{ccc}2x^2-1=y^2\\2x^2-1=3y^2\end{array}\right\ \ominus \ \left\{\begin{array}{l}2x^2-1=y^2\\0=-2y^2\end{array}\right\\\\\\\left\{\begin{array}{l}2x^2-1=0\\y^2=0\end{array}\right\ \ \left\{\begin{array}{l}x^2=\dfrac{1}{2}\\y=0\end{array}\right\ \ \left\{\begin{array}{l}x=\pm \dfrac{1}{\sqrt2}\\y=0\end{array}\right\ \ \ Otvet:\ \Big(-\dfrac{1}{\sqrt2}\, ;\, 0\Big)\ ,\ \Big(\dfrac{1}{\sqrt2}\, ;\, 0\Big)\ .$

$3)\ \ \left\{\begin{array}{l}x^2+y^2-10=0\\xy-3=0\end{array}\right\ \ \left\{\begin{array}{l}x^2+2xy+y^2=10+2xy\\xy=3\end{array}\right\ \ \left\{\begin{array}{l}(x+y)^2=10+2\cdot 3\\xy=3\end{array}\right\\\\\\\left\{\begin{array}{l}(x+y)^2=16\\xy=3\end{array}\right\ \ \left\{\begin{array}{l}x+y=\pm 4\\xy=3\end{array}\right\ \ \ \Rightarrow$

$a)\ \ \left\{\begin{array}{l}x+y=4\\xy=3\end{array}\right\ \ \left\{\begin{array}{l}y=4-x\\x(4-x)=3\end{array}\right\ \ \left\{\begin{array}{l}y=4-x\\x^2-4x+3=0\end{array}\right\ \ \left\{\begin{array}{l}y_1=3\ ,\ y_2=1\\x_1=1\ ,\ x_2=3\end{array}\right$

$b)\ \ \left\{\begin{array}{l}x+y=-4\\xy=3\end{array}\right\ \ \left\{\begin{array}{l}y=-4-x\\x\, (-4-x)=3\end{array}\right\ \ \left\{\begin{array}{l}y=-4-x\\x^2+4x+3=0\end{array}\right\ \ \left\{\begin{array}{l}y_1=-1\ ,\ y_2=-3\\x_1=-3\ ,\ x_2=-1\end{array}\right\\\\\\Otvet:\ \ (1,3)\ ,\ (3,1)\ ,\ (-3-1)\ ,\ (-1,-3)\ .$