§9. Арифметическкая прогрессия — 28. Формула суммы первых n членов арифметической прогрессии — 587 — стр. 166

$\{\begin{array}{l}9x^2+9y^2=13\\ 3xy=2\end{array}\iff \{\begin{array}{l}9x^2+9\cdot {\left(\frac{2}{3x}\right)}^2=13\\ y=\frac{2}{3x}\end{array}$

$9x^2+\frac{4}{x^2}=13\mid \cdot x^2$

$9x^4-13x^2+4=0\iff (x^2-1)(9x^2-4)=0\iff \{\begin{array}{l}{x}^2=\frac{4}{9}\\ x^2=1\end{array}$

$\{\begin{array}{l}\{\begin{array}{l}x=\pm \frac{2}{3}\\ x=\pm 1\end{array}\\ y=\frac{2}{3x}\end{array}\iff \{\begin{array}{l}\{\begin{array}{l}x=\frac{2}{3}\\ y=\frac{2}{3}\cdot (-\frac{3}{2})=-1\end{array}\\ \{\begin{array}{l}x=\frac{2}{3}\\ y=1\end{array}\\ \{\begin{array}{l}x=-1\\ y=-\frac{2}{3}\end{array}\\ \{\begin{array}{l}x=1\\ y=\frac{2}{3}\end{array}\end{array}$

Ответ: $(-\frac{2}{3};-1),(\frac{2}{3};1),(-1;-\frac{2}{3}),(1;\frac{2}{3})$.

$\{\begin{array}{l}{x}^2+y^2=29\\ y^2-4x^2=9\end{array}\iff \{\begin{array}{l}5x^2=29-9\\ y^2=29-x^2\end{array}\iff \{\begin{array}{l}{x}^2=4\\ y^2=29-x^2\end{array}\iff$

$\iff \{\begin{array}{l}{x}^2=4\\ y^2=25\end{array}\iff \{\begin{array}{l}x=\pm 2\\ y=\pm 5\end{array}\iff \{\begin{array}{l}\{\begin{array}{l}x=-2\\ y=-5\end{array}\\ \{\begin{array}{l}x=-2\\ y=5\end{array}\\ \{\begin{array}{l}x=2\\ y=-5\end{array}\\ \{\begin{array}{l}x=2\\ y=5\end{array}\end{array}$

Ответ: $(-2;-5),(-2;5),(2;-5),(2;5)$.

$\{\begin{array}{l}2x^2+xy=6\\ 3x^2+xy-x=6\end{array}\iff \{\begin{array}{l}{x}^2-x=6-6\\ 2x^2+xy=6\end{array}\iff \{\begin{array}{l}x(x-1)=0\\ 2x^2+xy=6\end{array}\iff$

$\iff \{\begin{array}{l}\{\begin{array}{l}x=0\\ x=1\end{array}\\ 2x^2+xy=6\end{array}\iff \{\begin{array}{l}\{\begin{array}{l}x=0\\ 2\cdot 0^2+0y=6\end{array}\\ \{\begin{array}{l}x=1\\ 2\cdot 1^2+y=6\end{array}\end{array}\iff \{\begin{array}{l}\{\begin{array}{l}x=0\\ y\in \mathrm{\varnothing }\end{array}\\ \{\begin{array}{l}x=1\\ y=4\end{array}\end{array}\iff \{\begin{array}{l}x=1\\ y=4\end{array}$

Ответ: $(1;4)$.

$\{\begin{array}{l}3x^2-2y^2=25\\ x^2-y^2+y=5|\cdot 3\end{array}\iff \{\begin{array}{l}3x^2-2y^2=25\\ 3x^2-3y^2+3y=15\end{array}\iff \{\begin{array}{l}{y}^2-3y=25-15\\ x^2=y^2-y+5\end{array}\iff$

$\iff \{\begin{array}{l}{y}^2-3y-10=0\\ x^2=y^2-y+5\end{array}\iff \{\begin{array}{l}(y+2)(y-5)=0\\ x^2=y^2-y+5\end{array}\iff \{\begin{array}{l}\{\begin{array}{l}y=-2\\ y=5\end{array}\\ x^2=y^2-y+5\end{array}\iff$

$\iff \{\begin{array}{l}\{\begin{array}{l}y=-2\\ x^2=(-2{)}^2-(-2)+5=13\end{array}\\ \{\begin{array}{l}y=5\\ x^2=5^2-5+5=25\end{array}\end{array}\iff \{\begin{array}{l}\{\begin{array}{l}x=\pm \sqrt{13}\\ y=-2\end{array}\\ \{\begin{array}{l}x=\pm 5\\ y=5\end{array}\end{array}\iff \{\begin{array}{l}\{\begin{array}{l}x=-\sqrt{13}\\ y=-2\end{array}\\ \{\begin{array}{l}x=\sqrt{13}\\ y=-2\end{array}\\ \{\begin{array}{l}x=-5\\ y=5\end{array}\\ \{\begin{array}{l}x=5\\ y=5\end{array}\end{array}$

Ответ: $(-\sqrt{13};-2),(\sqrt{13};-2),(-5;5),(5;5)$.