Алгебра, вопрос задал annakuliova0 , 1 год назад

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Ответил sangers1959

Пояснення:

$\displaystyle \\\frac{4xy+(x-y)^2}{1+\frac{y}{x} } =\frac{4xy+x^2-2xy+y^2}{\frac{x+y}{x} }=\frac{x^2+2xy+y^2}{\frac{x+y}{x} } =\\\\=\frac{x*(x+y)^2}{x+y} =x*(x+y).$

$\displaystyle \\\frac{1}{x} -\frac{1}{y} +\frac{x}{y^2} =\frac{y^2-xy+x^2}{xy^2}=\frac{x^2-xy+y^2}{xy^2} .$

$\displaystyle \\\frac{1}{x} +\frac{x^2}{y^3} =\frac{y^3+x^3}{xy^3}=\frac{(x+y)*(x^2-xy+y^2)}{xy^3}.$

$\displaystyle \\\frac{x^2-xy+y^2}{xy^2}:\frac{(x+y)*(x^2-xy+y^2)}{xy^3} =\frac{(x^2-xy+y^2)*xy^3}{xy^2*(x+y)*(x^2-xy+y^2)} }=\\\\ =\frac{y}{x+y} .$

$\displaystyle \\x*(x+y):\frac{y}{x+y} =\frac{x*(x+y)*(x+y)}{y} =\frac{x}{y} *(x+y)^2.$

$\displaystyle \\\frac{xy-1}{x-4y} =0 \ \ \ \ \ \ \Rightarrow\\\\\left \{ {{xy-1=0} \atop {x-4y\neq 0}} \right. \ \ \ \ \ \ \left \{ {{xy=1} \atop {x\neq 4y\ |:4}} \right. \ \ \ \ \ \ \left \{ {{y=\frac{1}{x} } \atop {y\neq \frac{x}{4} }} \right. .$

10.

$\displaystyle \\\left \{ {{\frac{3x}{10}-\frac{x}{6} \geq 4 \ |*30} \atop {\frac{x}{2}-\frac{2x}{3} \geq -8\ |*6 }} \right. \ \ \ \ \ \ \left \{ {{9x-5x\geq 120} \atop {3x-4x\geq -48}} \right. \ \ \ \ \ \left \{ {{4x\geq 120\ |:4} \atop {-x\geq -48\ |*(-1)}} \right.\ \ \ \ \left \{ {{x\geq 30} \atop {x\leq 48}} \right. .\\\\x\in[30;48].$