Question

Пожалуйста очень срочно! решите ЛИНЕЙНЫЕ СИСТЕМЫ! 91. 3) и 2). номер 92. 3)​. АЛГЕБРа
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Answer 1

$91.2)\ \ \left\{\begin{array}{l}x^2+3y^2-4x-5y-8=0\\x-y+1=0\end{array}\right\ \ \left\{\begin{array}{l}(y-1)^2+3y^2-4(y-1)-5y-8=0\\x=y-1\end{array}\right\\\\\\\left\{\begin{array}{ccc}4y^2-11y-3=0\\x=y-1\end{array}\right\ \ \left\{\begin{array}{l}D=169\ ,\ y_1=-\frac{1}{4}\ ,\ y_2=3\\x=y-1\end{array}\right\ \ \left\{\begin{array}{ccc}y_1=-0,25\ ,\ y_2=3\\x_1=-1,25\ ,\ x_2=2\end{array}\right\\\\\\Otvet:\ \ (-1,25\, ;\, -0,25)\ ,\ (\, 2\, ;\, 3\, )\ .$

$91.3)\ \ \left\{\begin{array}{l}4x^2+5y^2=16\\x^2+5y^2=25\end{array}\right\ \ominus \ \left\{\begin{array}{l}3x^2=-9\\x^2+5y^2=25\end{array}\right\ \ \Rightarrow \ \ \left\{\begin{array}{l}x\in \varnothing \\x^2+5y^2=25\end{array}\right$

Ответ:  нет решений .

$92.3)\ \ \left\{\begin{array}{l}x^2+y^2=65\\xy=28\end{array}\right\ \ \left\{\begin{array}{l}(x+y)^2-2xy=65\\xy=28\end{array}\right\ \ \left\{\begin{array}{l}(x+y)^2-2\cdot 28=65\\xy=28\end{array}\right\\\\\\\left\{\begin{array}{ccc}(x+y)^2=121\\xy=28\end{array}\right\ \ \left\{\begin{array}{l}x+y=\pm 11\\xy=28\end{array}\right$

$a)\ \ \left\{\begin{array}{l}x+y=-11\\xy=28\end{array}\right\ \ \ \ \qquad \qquad b)\ \ \left\{\begin{array}{l}x+y=11\\xy=28\end{array}\right\\\\\\\left\{\begin{array}{l}x=-y-11\\(-y-11)\cdot y=28\end{array}\right\ \ \ \ \qquad \qquad \left\{\begin{array}{l}x=-y+11\\(-y+11)\cdot y=28\end{array}\right$

$\left\{\begin{array}{l}x=-y-11\\y^2+11y+28=0\end{array}\right\ \ \ \qquad \qquad \ \left\{\begin{array}{l}x=-y+11\\y^2-11y+28=0\end{array}\right\\\\\\\left\{\begin{array}{ccc}x_1=-4\ ,\ x_2=-7\\y_1=-7\ ,\ y_2=-4\end{array}\right\ \ \ \qquad \qquad \left\{\begin{array}{ccc}x_1=7\ ,\ x_2=4\\y_1=4\ ,\ y_2=7\end{array}\right$

$Otvet:\ \ (-4;-7)\ ,\ (-7;-4)\ ,\ (7;4)\ ,\ (4;7)\ .$

Answer 2

Ответ:

2)(2;3)

(-1 1/4; -1/4)

3) нет решений

3)(-7;-4)

(-4;-7)

(4; 7)

(7; 4)

Объяснение: