Question

Решите систему уравнений:

Answer 1

Ответ:

$(-\sqrt{23} ;-3\sqrt{23} ), (\sqrt{23} ;3\sqrt{23} ), \left(-\dfrac{7}{\sqrt{3} } ;-\dfrac{1}{\sqrt{3} } \right), \left(\dfrac{7}{\sqrt{3} } ;\dfrac{1}{\sqrt{3} } \right).$

Объяснение:

Решить систему уравнений

$\left \{\begin{array}{l} \dfrac{2x+y}{x-2y}-\dfrac{3(x-2y)}{2x+y} =2, \\\\ x^{2} +3xy-y^{2} = 23. \end{array} \right.$

Рассмотрим первое уравнение системы

$\dfrac{2x+y}{x-2y}-\dfrac{3(x-2y)}{2x+y} =2.$

Пусть $\dfrac{2x+y}{x-2y}=t$  , а    $\dfrac{x-2y}{2x+y} =\dfrac{1}{t}$.

Тогда уравнение принимает вид:

$t-\dfrac{3}{t} =2|\cdot t\neq 0;\\\\t^{2}-2t-3=0;\\\\D= (-2)^{2} -4\cdot1\cdot(-3)=4+12=16=4^{2} ;\\\\t{_1}= \dfrac{2-4}{2} =-\dfrac{2}{2} =-1;\\\\t{_1}= \dfrac{2+4}{2} =\dfrac{6}{2} =3$

Тогда получим две системы

$1)\left \{\begin{array}{l} \dfrac{2x+y}{x-2y} =-1, \\\\ x^{2} +3xy-y^{2} = 23. \end{array} \right.\Leftrightarrow\left \{\begin{array}{l}2x+y=-x+2y, \\\\ x^{2} +3xy-y^{2} = 23. \end{array} \right.\Leftrightarrow\left \{\begin{array}{l} y=3x, \\\\ x^{2} +3xy-y^{2} = 23. \end{array} \right.\Leftrightarrow$

$\Leftrightarrow\left \{\begin{array}{l} y=3x, \\\\ x^{2} +3x\cdot(3x)-(3x)^{2} = 23. \end{array} \right.\Leftrightarrow\left \{\begin{array}{l} y=3x, \\\\ x^{2} +9x^{2} -9x^{2} = 23. \end{array} \right.\Leftrightarrow\left \{\begin{array}{l} y=3x, \\\\ x^{2} = 23. \end{array} \right.\Leftrightarrow$

$\Leftrightarrow\left \{\begin{array}{l} y=3x, \\\\ \left [\begin{array}{l} x=-\sqrt{23} \\ x = \sqrt{23} \end{array} \right. \end{array} \right.\Leftrightarrow \left [\begin{array}{l} \left \{\begin{array}{l} x = -\sqrt{23} \\ y = -3\sqrt{23} \end{array} \right. \\\\ \left \{\begin{array}{l} x =\sqrt{23} \\ y = 3\sqrt{23} \end{array} \right.\end{array} \right.$

$2)\left \{\begin{array}{l} \dfrac{2x+y}{x-2y} =3, \\\\ x^{2} +3xy-y^{2} = 23. \end{array} \right.\Leftrightarrow\left \{\begin{array}{l}2x+y=3x-6y, \\\\ x^{2} +3xy-y^{2} = 23. \end{array} \right.\Leftrightarrow\left \{\begin{array}{l} x=7y, \\\\ x^{2} +3xy-y^{2} = 23. \end{array} \right.\Leftrightarrow$

$\Leftrightarrow\left \{\begin{array}{l} x=7y, \\\\ (7y)^{2} +3y\cdot (7y)-y^{2} = 23. \end{array} \right.\Leftrightarrow\left \{\begin{array}{l} x=7y, \\\\ 49y^{2} +21y^{2} -y^{2} = 23. \end{array} \right.\Leftrightarrow\left \{\begin{array}{l} x=7y, \\\\ 69y^{2} = 23. \end{array} \right.\Leftrightarrow$

$\Leftrightarrow\left \{\begin{array}{l} x=7y, \\\\ \left [\begin{array}{l} y=-\dfrac{1}{\sqrt{3} } \\ y =\dfrac{1}{\sqrt{3} } \end{array} \right. \end{array} \right.\Leftrightarrow \left [\begin{array}{l} \left \{\begin{array}{l} x = -\dfrac{7}{\sqrt{3} } \\ y = -\dfrac{1}{\sqrt{3} } \end{array} \right. \\\\ \left \{\begin{array}{l} x =\dfrac{7}{\sqrt{3} } \\ y = \dfrac{1}{\sqrt{3} } \end{array} \right.\end{array} \right.$

Значит, заданная система имеет 4 решения

$(-\sqrt{23} ;-3\sqrt{23} ), (\sqrt{23} ;3\sqrt{23} ), \left(-\dfrac{7}{\sqrt{3} } ;-\dfrac{1}{\sqrt{3} } \right), \left(\dfrac{7}{\sqrt{3} } ;\dfrac{1}{\sqrt{3} } \right).$

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