Question

x^2+xy^3 = 192
x^3y+x^2y^2 = 96

Answer 1

Ответ:

$\begin{cases} {x}^{2} {y}^{2} + x {y}^{3} = 192 \\ {x}^{3} y + {x}^{2} {y}^{2} = 96 \end{cases}$

$\begin{cases}x {y}^{2} (x + y) = 192 \\ {x}^{2} y(x + y) = 96 \end{cases}$

$\frac{x {y}^{2} (x + y)}{ {x}^{2}y(x + y) } = \frac{192}{96}$

сокращаем на общий делитель x+y;y;x и 96:

$\frac{y}{x} = 2$

$x \times \frac{y}{x} = 2x$

сокращаем на общий делитель x:

$y = 2x$

${x}^{3} \times 2x + {x}^{2} \times {(2x)}^{2} = 96$

${x}^{3} \times 2x + {x}^{2} \times 4 {x}^{2} = 96$

$(1 + 2) \times 2 {x}^{2} \times x \times x = 96$

$3 \times 2 {x}^{2} \times x \times x = 96$

$6 {x}^{4} = 96$

делим обе стороны уравнения на 6:

${x}^{4} = 16$

$x = ±2$

$x = - 2 \\ x = 2$

${( - 2)}^{2} \times {y}^{2} + {( - 2)y}^{3} = 192 \\ {2}^{2} {y}^{2} + 2 {y}^{3} = 192$

${( - 2)}^{2} \times {y}^{2} + ( - 2) {y}^{3} = 192 \\ {2}^{2} {y}^{2} - 2 {y}^{3} = 192 \\ 4 {y}^{2} - 2 {y}^{3} = 192 \\ \div 2 \\ 2 {y}^{2} - {y}^{3} = 96 \\ 2 {y}^{2} - {y}^{3} - 96 = 0 \\ - {y}^{3} + 2 {y}^{2} - 96 = 0 \\ - {y}^{3} - 4 {y}^{2} + 6 {y}^{2} - 96 = 0 \\ - {y}^{2} (y + 4) + 6( {y}^{2} - 16) = 0 \\ - {y}^{2} (y + 4) + 6(y - 4)(y + 4) = 0 \\ - (y + 4)( {y}^{2} - 6(y - 4)) = 0 \\ - (y + 4)( {y}^{2} - 6y + 24) = 0 \\ (y + 4)( {y}^{2} - 6y + 24) = 0 \\ y + 4 = 0 \\ {y}^{2} - 6y + 24 = 0 \\ y = - 4 \\ y∉ℝ \\ y = - 4$

${2}^{2} {y}^{2} + 2 {y}^{3} = 192 \\ 4 {y}^{2} + 2 {y}^{3} = 192 \\ \div 2 \\ 2 {y}^{2} + {y}^{3} = 96 \\ 2 {y}^{2} + {y}^{3} - 96 = 0 \\ {y}^{3} + 2 {y}^{2} - 96 = 0 \\ {y}^{3} - 4 {y}^{2} + 6 {y}^{2} - 96 = 0 \\ {y}^{2} (y - 4) + 6( {y}^{2} - 16) = 0 \\ {y}^{2} (y - 4) + 6(y - 4)(y + 4) = 0 \\ (y - 4)( {y}^{2} + 6(y + 4)) = 0 \\ (y - 4)( {y}^{2} + 6y + 24) = 0 \\ y - 4 = 0 \\ {y}^{2} + 6y + 24 = 0 \\ y = 4 \\ y∉ℝ \\ y = 4$

$y = - 4 \\ y = 4$

$(x_{1}, y_{1})=(-2, -4) \\ (x_{2}, y_{2})=(2,4)$

$\begin{cases} {( - 2)}^{2} \times {( - 4)}^{2} + ( - 2) \times {( - 4)}^{3} = 192 \\ {( - 2)}^{3} \times ( - 4) + {( - 2)}^{2} \times {( - 4)}^{2} = 96 \end{cases} \\ \begin{cases} {2}^{2} \times {4}^{2} + 2 \times {4}^{3} = 192 \\ {2}^{3} \times 4 + {2}^{2} \times {4}^{2} = 96 \end{cases}$

$\begin{cases} 192 = 192 \\ 96 = 96\end{cases} \\ \begin{cases} 192 = 192 \\ 96 = 96\end{cases}$

$(x_{1}, y_{1})=(-2, -4) \\ (x_{2}, y_{2})=(2, 4)$