tex]...

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5) Область определения [tex]y = \frac{ \sqrt{3x - x^{2} }}{2 - x} [/tex]
7) Решите систему уравнений [tex] \left \{ {{3xy + x = 10} \atop {3xy + y = 12}} \right. [/tex]

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hote

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1) Область определения
$\displaystyle y= \frac{ \sqrt{3x-x^2}}{2-x}\\\\ \left \{ {{2-x \neq 0} \atop {3x-x^2 \geq 0}} \right. \\\\ \left \{ {{x \neq 2} \atop {x(3-x) \geq 0}} \right.\\\\ \left \{ {{x \neq 2} \atop {x\in [0;3]}} \right.\\\\ODZ: [0;2) (2;3]$

2) Решить систему
$\displaystyle \left \{ {{3xy+x=10} \atop {3xy+y=12}} \right.$

вычтем из второго уравнения первое
$\displaystyle y-x=2\\\\y=2+x$

подставим в первое

$\displaystyle 3x(2+x)+x=10\\\\6x+3x^2+x-10=0\\\\3x^2+7x-10=0\\\\D=49+120=169=13^2\\\\x_{1.2}= \frac{-7 \pm 13}{6}\\\\x_1=1; x_2=-10/3\\\\y_1=3; y_2=-4/3$

Ответ (1;3) и (-10/3; -4/3)

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