Question

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Answer 1

$\left \{ {{x^2+3xy=18} \atop {xy+4y^2=7}} \right. \\\\3xy=18-x^2\\y=\frac{18-x^2}{3x};\quad \:x\ne \:0\\\\\left \{ {{y=\frac{18-x^2}{3x};\quad} \atop {xy+4y^2=7}} \right. \\\\\left \{ {{y=\frac{18-x^2}{3x};\quad} \atop {x\frac{18-x^2}{3x}+4\left(\frac{18-x^2}{3x}\right)^2=7}} \right. \\\\\frac{18-x^2}{3}+\frac{4\left(18-x^2\right)^2}{9x^2}=7\\\\\frac{18-x^2}{3}\cdot \:9x^2+\frac{4\left(18-x^2\right)^2}{9x^2}\cdot \:9x^2=7\cdot \:9x^2\\\\3x^2\left(-x^2+18\right)+4\left(18-x^2\right)^2=63x^2$

$x^4-90x^2+1296=63x^2\\x^4-153x^2+1296=0\\| \:\: u=x^2\mathrm{\:and\:}u^2=x^4\\u^2-153u+1296=0\\\\\u_{1,\:2}=\frac{-\left(-153\right)\pm \sqrt{\left(-153\right)^2-4\cdot \:1\cdot \:1296}}{2\cdot \:1}\\\\u_{1}=\frac{\sqrt{23409-5184}}{2\cdot \:1} =\frac{153+\sqrt{18225}}{2}=\frac{153+135}{2}=144\\\\u_{2}=\frac{153-135}{2}=\frac{18}{2}=9\\u_{1}=144,\:u_{2}=9\\| \: u=x^2\\x^2=144\\x_{1}=\sqrt{144},\:x_{2}=-\sqrt{144}\\x_{1}=12,\:x_{2}=-12$

$x^4-90x^2+1296=63x^2\\x^4-153x^2+1296=0\\| \:\: u=x^2\mathrm{\:and\:}u^2=x^4\\u^2-153u+1296=0\\\\\u_{1,\:2}=\frac{-(-153)\pm \sqrt{(-153)^2-4\cdot \:1\cdot \:1296}}{2\cdot \:1}\\\\u_{1}=\frac{\sqrt{23409-5184}}{2\cdot \:1} =\frac{153+\sqrt{18225}}{2}=\frac{153+135}{2}=144\\\\u_{2}=\frac{153-135}{2}=\frac{18}{2}=9\\u_{1}=144,\:u_{2}=9\\| \: u=x^2\\\\x^2=144\\x_{1}=\sqrt{144},\:x_{2}=-\sqrt{144}\\x_{1}=12,\:x_{2}=-12\\\\x^2=9\\x_{3}=\sqrt{9},\:x_{4}=-\sqrt{9}\\x_{3}=3,\:x_{4}=-3$

$x_{1}=12,\:x_{2}=-12,\:x_{3}=3,\:x_{4}=-3\\\\x^2+3xy=18 ,\:\: x=12\\12^2+3\cdot \:12y=18\\144+36y=18\\36y=-126\\\frac{36y}{36}=\frac{-126}{36}\\y=-\frac{7}{2}\\\\x^2+3xy=18 ,\:\: x=-12\\\left(-12\right)^2+3\left(-12\right)y=18\\144-36y=18\\-36y=-126\\y=\frac{7}{2}\\\\x^2+3xy=18 ,\:\: x=3\\3^2+3\cdot \:3y=18\\9+9y=18\\9y=9\\y=1$

$x^2+3xy=18 ,\:\: x=-3\\(-3)^2+3(-3)y=18\\9-9y=18\\-9y=9\\y=-1$

Ответ:

$\begin{pmatrix}x=12,\:&y=-\frac{7}{2}\\ x=-12,\:&y=\frac{7}{2}\\ x=3,\:&y=1\\ x=-3,\:&y=-1\end{pmatrix}$