ОГЭ
Задание 6068
Решите систему уравнений: $$\left\{\begin{matrix}5(2x-1)+1=6(y+1)-8 & & \\2(x+3y)+5=3(y-2x)+4 & &\end{matrix}\right.$$
$$\left\{\begin{matrix}5(2x-1)+1=6(y+1)-8\\2(x+3y)+5=3(y-2x)+4\end{matrix}\right.\Leftrightarrow $$$$\left\{\begin{matrix}10x-5+1-6y-6+8=0\\2x+6y+5-3y+6x-4=0\end{matrix}\right.\Leftrightarrow $$$$\left\{\begin{matrix}10x-6y-2=0\\8x+3y+1=0 \end{matrix}\right.\Leftrightarrow $$$$\left\{\begin{matrix}10x-6y-2=0\\16x+6y+2=0\end{matrix}\right.$$ Сложим первое и второе , $$10x+16x-6y+6y-2+2=0$$ $$26x=0 \Rightarrow x=0$$ Тогда : $$10*0-6y-2=0 \Leftrightarrow 6y=-2 \Leftrightarrow y=-\frac{1}{3}$$
Задание 6306
Решите систему уравнений $$\left\{\begin{matrix}x^{2}+xy+y^{2}=37\\ x^{3}-y^{3}=37\end{matrix}\right.$$
$$\left\{\begin{matrix}x^{2}+xy+y^{2}=37\\x^{3}-y^{3}=37\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}x^{2}+xy+y^{2}=37\\(x-y)(x^{2}+xy+y^{2})=37\end{matrix}\right.$$ Поделим второе на первое уравнение :$$x-y=1\Leftrightarrow x=1+y$$ $$(1+y)^{2}+(1+y)y+y^{2}=37$$ $$1+2y+y^{2}+y+y^{2}+y^{2}=37$$ $$3y^{2}+3y-36=0|:3$$ $$y^{2}+y-12=0\Leftrightarrow$$ $$D=1+48=49\Leftrightarrow$$ $$\left\{\begin{matrix}y_{1}=\frac{-1+7}{2}=3\\y_{2}=\frac{-1-7}{2}=-4\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}x_{1}=1+3=4\\x_{2}=1-4=-3\end{matrix}\right.$$
Задание 6502
Решите систему уравнений: $$\left\{\begin{matrix}(x+y)^{2}+2x=35-2y\\ (x-y)^{2}-2y=3-2x\end{matrix}\right.$$
$$\left\{\begin{matrix}(x+y)^{2}+2x=35-2y\\(x-y)^{2}-2y=3-2x\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}(x+y)^{2}=35-2(x+y)\\(x-y)^{2}=3-2(x-y)\end{matrix}\right.$$
Пусть x+y=a; x-y=6
$$\left\{\begin{matrix}a^{2}=35-2a\\b^{2}=3-2b\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}a^{2}+2a-35=0\\b^{2}+2b-3=0\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}\left[\begin{matrix}a=-7\\a=5\end{matrix}\right.\\\left[\begin{matrix}b=-3\\b=1\end{matrix}\right.\end{matrix}\right.$$
Получаем четыре пары решений: (-7;-3);(-7;1);( 5;-3); (5;1)
1) $$\left\{\begin{matrix}x+y=-7\\x-y=-3\end{matrix}\right.\Leftrightarrow$$ $$2x=-10\Leftrightarrow$$ $$x=-5\Leftrightarrow$$ $$y=-2$$
2) $$\left\{\begin{matrix}x+y=-1\\x-y=1\end{matrix}\right.\Leftrightarrow$$ $$2x=-6\Leftrightarrow$$$$x=-3\Leftrightarrow$$ $$y=-4$$
3) $$\left\{\begin{matrix}x+y=5\\x-y=-3\end{matrix}\right.\Leftrightarrow$$ $$2x=2\Leftrightarrow$$ $$x=1\Leftrightarrow$$ $$y=4$$
4) $$\left\{\begin{matrix}x+y=5\\x=y=1\end{matrix}\right.\Leftrightarrow$$ $$2x=6\Leftrightarrow$$ $$x=3\Rightarrow$$ $$y=2$$
Задание 6549
Решите систему уравнений: $$\left\{\begin{matrix}x^{2}-xy+y^{2}=79\\ x-y=7\end{matrix}\right.$$
$$\left\{\begin{matrix}x^{2}-xy+y^{2}=79\\x-y=7\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}(7+y)^{2}-(7+y)y+y^{2}=79\\x=7+y\end{matrix}\right.$$
$$49+14y+y^{2}-7y-y^{2}+y^{2}-79=0\Leftrightarrow$$$$y^{2}+7y-30=0$$
$$\left\{\begin{matrix}y_{1}+y_{2}=-7\\y_{1}y_{2}=-30\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}y_{1}=-10\\y_{2}=3\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}x_{1}=7-10=-3\\x_{2}=7+3-10\end{matrix}\right.$$
Задание 6596
Решите систему уравнений $$\left\{\begin{matrix}xy+x-y=7\\x^{2}y-xy^{2}=6\end{matrix}\right.$$
$$\left\{\begin{matrix}xy+x-y=7\\x^{2}y-xy^{2}=6\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}xy+(x-y)=7\\xy(x-y)=6\end{matrix}\right.$$
Пусть xy=a; x-y=b.
$$\left\{\begin{matrix}a+b=7\\ab=6\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}\left\{\begin{matrix}a=1\\b=6\end{matrix}\right. (1)\\\left\{\begin{matrix}a=6\\b=1\end{matrix}\right. (2)\end{matrix}\right.$$
1) $$\left\{\begin{matrix}xy=1\\x-y=6\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}6y+y^{2}=1\\x=6+y\end{matrix}\right.$$
$$y^{2}+6y-1=0$$, $$D=36+4=40\Leftrightarrow$$ $$\left[\begin{matrix}y_{1}=\frac{-6+\sqrt{40}}{2}=-3+\sqrt{10}\\y_{2}=\frac{-6-\sqrt{40}}{2}=-3-\sqrt{10}\end{matrix}\right.\Leftrightarrow$$ $$\left[\begin{matrix}x_{1}=3+\sqrt{10}\\x_{2}=3-\sqrt{10}\end{matrix}\right.$$
2)$$\left\{\begin{matrix}xy=6\\x-y=1\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}y+y^{2}-6=0\\x=1+y\end{matrix}\right.$$
$$y^{2}+y-6=0\Leftrightarrow$$ $$\left\{\begin{matrix}y_{1}+y_{2}=-1\\y_{1}y_{2}=-6\end{matrix}\right.\Leftrightarrow$$ $$\left[\begin{matrix}y_{1}=-3\\y_{2}=2\end{matrix}\right.\Leftrightarrow$$ $$\left[\begin{matrix}x_{1}=-2\\x_{2}=3\end{matrix}\right.$$
Задание 6711
Решите систему уравнений $$\left\{\begin{matrix}x+xy+y=5\\ x^{2}+xy+y^{2}=7\end{matrix}\right.$$
$$\left\{\begin{matrix}x+xy+y=5\\x^{2}+xy+y^{2}=7\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}(x+y)+xy=5\\x^{2}+2xy+y^{2}-xy=7\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}(x+y)+xy=5\\(x+y)^{2}-xy=7\end{matrix}\right.$$
Пусть x+y=a; xy=b
$$\left\{\begin{matrix}a+b=5(1)\\a^{2}-b=7(2)\end{matrix}\right.\Leftrightarrow$$ $$b=5-a$$
Сложим (1) и (2): $$a^{2}+a=12\Leftrightarrow$$ $$a^{2}+a-12=0$$
$$\left\{\begin{matrix}a_{1}+a_{2}=-1\\a_{1}*a_{2}=-12\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}a_{1}=-4\\a_{2}=3\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}b=5-(-4)=9\\b=5-3=2\end{matrix}\right.$$
$$\left[\begin{matrix}\left\{\begin{matrix}x+y=-4\\xy=9\end{matrix}\right.\\\left\{\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow$$ $$\left[\begin{matrix}\left\{\begin{matrix}x=4-y\\(-4-y)y=9\end{matrix}\right. (1)\\\left\{\begin{matrix}x=3+y\\(3-y)y=2\end{matrix}\right.(2)\end{matrix}\right.$$
(1): $$-y^{2}-4y-9=0\Leftrightarrow$$ $$y^{2}+4y+9=0\Leftrightarrow$$ $$D=16-36<0\Rightarrow$$ решений нет
(2): $$3y-y^{2}=2\Leftrightarrow$$ $$y^{2}-3y+2=0\Leftrightarrow$$ $$\left[\begin{matrix}y_{1}=1\\y_{2}=2\end{matrix}\right.\Leftrightarrow$$ $$\left[\begin{matrix}x_{1}=2\\x_{2}=1\end{matrix}\right.$$
Задание 6904
Решите систему уравнений $$\left\{\begin{matrix} x+4y=18\\x^{2}+y^{2}=20\end{matrix}\right.$$
$$\left\{\begin{matrix}x+4y=18\\x^{2}+y^{2}=20\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}x=18-4y\\(18-4y)^{2}+y^{2}=20\end{matrix}\right.$$
$$324-144y+16y^{2}+y^{2}-20=0\Leftrightarrow$$$$17y^{2}-144y+304=0$$
$$D=20736-20672=64$$
$$y_{1}=\frac{144+8}{34}=\frac{76}{77}\Rightarrow$$ $$x_{1}=18-4*\frac{76}{77}=\frac{2}{17}$$
$$y_{2}=\frac{144-8}{34}=4\Rightarrow$$ $$ x_{2}=18-4*4=2$$
Задание 7160
Решите систему уравнений $$\left\{\begin{matrix}(x-1)(y-1)=1\\x^2y+xy^2=16 \end{matrix}\right.$$
$$\left\{\begin{matrix}(x-1)(y-1)=1\\x^{2}y+xy^{2}=16\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}xy-x*y+1=1\\xy(x+y)=16\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}xy-(x+y)=0\\xy(x+y)=16\end{matrix}\right.$$
Пусть: $$xy=a$$ , $$x+y=b$$
$$\left\{\begin{matrix}x-b=0\\ab=16\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}x=b\\a^{2}=16\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}b=\pm 4\\a=\pm 4\end{matrix}\right.\Leftrightarrow$$ $$\left[\begin{matrix}\left\{\begin{matrix}xy=4\\x+y=4\end{matrix}\right.\\\left\{\begin{matrix}xy=-4\\x+y=-4\end{matrix}\right.\end{matrix}\right.\Leftrightarrow$$ $$\left[\begin{matrix}\left\{\begin{matrix}4y-y^{2}-4=0\\x=4-y\end{matrix}\right.\\\left\{\begin{matrix}-4y-y^{2}+4=0\\x=-4-y\end{matrix}\right.\end{matrix}\right. \Leftrightarrow$$ $$\left[\begin{matrix}\left\{\begin{matrix}y^{2}-4y+4=0\\x=4-y\end{matrix}\right.\\\left\{\begin{matrix}y^{2}+4y-4=0\\x=-4-y\end{matrix}\right.\end{matrix}\right.\Leftrightarrow$$ $$\left[\begin{matrix}\left\{\begin{matrix}y=2\\x=2\end{matrix}\right.\\\left\{\begin{matrix}y=-2+\sqrt{2}\\x=-2-\sqrt{2}\end{matrix}\right.\\\left\{\begin{matrix}y=-2-\sqrt{2}\\x=-2+\sqrt{2}\end{matrix}\right.\end{matrix}\right.$$
$$y^{2}+4y-4=0$$
$$D=16+16=32$$
$$y_{1,2}=\frac{-4\pm \sqrt{32}}{2}=-2\pm \sqrt{2}$$