✨라니수행평가공부해✨

고라니공부해

고라니수학해

고라니제발수행평가던지지마

(대충 친구 수학 알려주려고 만든 것입니다)

(+) 컴퓨터로 보면 매우 구조적이고 예쁘게 잘 썼는데 휴대폰으로 보면 순서가..바뀐다ㅠㅠㅠㅠㅠ

$\begin{cases} x^2-xy-2y^2=0 \\ 2x^2+y^2 = 9 \end{cases}$

$(x-2y)(x+y)=0$

$\Rightarrow$ $x=2y$ or $x=-y$

$1)\;x=2y\\\;2(2y)^2+y^2=9\\\;\Rightarrow9y^2=9 \Rightarrow y=\pm1 \\\;\Rightarrow (2,1),(-2,-1)$

$2)\;x=-y\\\;2(-y)^2+y^2=9\\\;\Rightarrow 3y^2=9\Rightarrow y=\pm\sqrt{3}\\\;\Rightarrow (-\sqrt{3},\sqrt{3}),(\sqrt{3},-\sqrt{3})$

$\begin{cases} x^2+xy-2y^2=0 \\ x^2+2xy+y^2 = 4 \end{cases}$

$(x+2y)(x-y)=0$

$\Rightarrow$$x=-2y$ or $x=y$

$(x+y)^2=4$

$1)\;x=-2y\\\;(-2y+y)^2=4\\\;\Rightarrow\; y^2=4\Rightarrow y=\pm2 \\\;\Rightarrow (-4,2),(4,-2)$

$2)\;x=y\\\;(y+y)^2=4\\\;\Rightarrow 4y^2=4\Rightarrow y=\pm1\\\;\Rightarrow (1,1),(-1,-1)$

$\begin{cases} x^2-2xy-3y^2=0 \\ x^2+y^2 = 40 \end{cases}$

$(x-3y)(x+y)=0$

$\Rightarrow$$x=3y$ or $x=-y$

$1)\;x=3y\\\;(3y)^2+y^2=40\\\;\Rightarrow10y^2=40 \Rightarrow y=\pm2 \\\;\Rightarrow (6,2),(-6,-2)$

$2)\;x=-y\\\;(-y)^2+y^2=40\\\;\Rightarrow 2y^2=40\Rightarrow y=\pm2\sqrt{5}\\\;\Rightarrow (-2\sqrt{5},2\sqrt{5}),(2\sqrt{5},-2\sqrt{5})$

$\begin{cases} 2x^2+3xy-2y^2=0 \\ x^2+xy = 12 \end{cases}$

$(2x-y)(x+2y)=0$

$\Rightarrow$ $x=\frac{1}{2}y$ or $x=-2y$

$x(x+y)=12$

$1)\;x=\frac{1}{2}y\\\;\frac{1}{2}y(\frac{1}{2}y+y)=12\\\;\Rightarrow\; \frac{3}{4}y^2=12\Rightarrow y^2=16\Rightarrow y=\pm4 \\\;\Rightarrow (2,4),(-2,-4)$

$2)\;x=-2y\\\;-2y(-2y+y)=12\\\;\Rightarrow 2y^2=12\Rightarrow y=\pm\sqrt{6}\\\;\Rightarrow (-2\sqrt{6},\sqrt{6}),(2\sqrt{6},-\sqrt{6})$

$\begin{cases} 2x^2+xy-y^2=0 \\ x^2+xy+y^2 = 7 \end{cases}$

$(2x-y)(x+y)=0$

$\Rightarrow$ $x=\frac{1}{2}y$ or $x=-y$

$1)\;x=\frac{1}{2}y\\\;(\frac{1}{2}y)^2+\frac{1}{2}y^2+y^2=7\\\;\Rightarrow\; \frac{7}{4}y^2=7\Rightarrow y=\pm2 \\\;\Rightarrow (1,2),(-1,-2)$

$2)\;x=-y\\\;(-y)^2+(-y)y+y^2=7\\\;\Rightarrow y^2=7\Rightarrow y=\pm\sqrt{7}\\\;\Rightarrow (\sqrt{7},-\sqrt{7}),(-\sqrt{7},\sqrt{7})$