Question

С решением ток.Плииииииз

Answer 1

$1)\; \; \left\{\begin{array}{l}x+xy+y=5\\x-xy+y=1\end{array}\right\; \; t=x+y\; ,\; \; p=xy\; \; \Rightarrow \; \;\left\{\begin{array}{l}t+p=5\\t-p=1\end{array}\right\; \; \oplus \\\\\\\left\{\begin{array}{l}2t=6\\p=t-1\end{array}\right\; \; \left\{\begin{array}{ccc}t=3\\p=2\end{array}\right\; \; \left\{\begin{array}{l}x+y=3\\xy=2\end{array}\right\; \; \left\{\begin{array}{l}y=3-x\\x(3-x)=2\end{array}\right\; \; \left\{\begin{array}{l}y=3-x\\x^2-3x+2=0\end{array}\right$

$\left\{\begin{array}{ccc}y_1=2\; ,\; y_2=1\\x_1=1\; ,\; x_2=2\end{array}\right\; \; \; \Rightarrow \; \; \; Otvet:\; (1;2)\; ,\; (2;1)$

$2)\; \; \left\{\begin{array}{l}xy-x-y=-7\\xy(x+y)=-6\end{array}\right\; \; t=x+y\; ,\; \; p=xy\; \; \left\{\begin{array}{l}p-t=-7\\pt=-6\end{array}\right\\\\\\\left\{\begin{array}{l}t=p+7\\p\, (p+7)=-6\end{array}\right\; \; \left\{\begin{array}{l}t=p+7\\p^2+7p+6=0\end{array}\right\; \; \left\{\begin{array}{l}t_1=1\; ,\; t_2=6\\p_1=-6\; ,\; p_2=-1\end{array}\right$

$a)\; \; \left\{\begin{array}{l}x+y=1\\xy=-6\end{array}\right\; \; \left\{\begin{array}{l}y=1-x\\x(1-x)=-6\end{array}\right\; \; \left\{\begin{array}{l}y=1-x\\x^2-x-6=0\end{array}\right\\\\\\\left\{\begin{array}{l}y_1=3\; ,\; y_2=-2\\x_1=-2\; ,\; x_2=3\end{array}\right\; \; \; \; \; \; \; \underline {(-2;3)\; ,\; (3;-2)\; }$

$b)\; \; \left\{\begin{array}{l}x+y=6\\xy=-1\end{array}\right\; \; \left\{\begin{array}{l}y=6-x\\x(6-x)=-1\end{array}\right\; \; \left\{\begin{array}{l}y=6-x\\x^2-6x-1=0\end{array}\right\\\\\\D/4=3^2-(-1)=10\; ,\; \; x_1=3-\sqrt{10}\; ,\; \; x_2=3+\sqrt{10}\\\\y_1=3+\sqrt{10}\; \; ,\; \; y_2=3-\sqrt{10}\\\\\underline {(3-\sqrt{10}\, ;\, 3+\sqrt{10}\; )\; \; ,\; \; (3+\sqrt{10}\, ;\, 3-\sqrt{10})\; }$

$3)\; \; \left\{\begin{array}{ccc}16x^2+8xy+y^2=36\\3x-y=8\end{array}\right\; \; \left\{\begin{array}{l}(4x+y)^2=36\\y=8-3x\end{array}\right\; \; \left\{\begin{array}{l}4x+y=\pm 6\\y=3x-8\end{array}\right\\\\\\a)\; \; \left\{\begin{array}{l}4x+y=6\\y=3x-8\end{array}\right\; \; \left\{\begin{array}{ccc}7x=14\\y=3x-8\end{array}\right\; \; \left\{\begin{array}{l}x=2\\y=-2\end{array}\; \; \; \; (2;-2)$

$b)\; \; \left\{\begin{array}{ccc}4x+y=-6\\y=3x-8\end{array}\right\; \; \left\{\begin{array}{ccc}7x=2\\y=3x-8\end{array}\right\; \; \left\{\begin{array}{ccc}x=\frac{2}{7}\\y=-\frac{50}{7}\end{array}\right\; \; \; \; (\frac{2}{7}\, ;\, -\frac{50}{7}\, )\\\\\\Otvet:\; \; (2;-2)\; \; ,\; \; (\frac{2}{7}\, ;\, -\frac{50}{7}\, )\; .$