Question

X^2(x+1)^2+(y-2)^2=0

Answer 1

$x^2(x+1)^2+(y-2)^2=0 \\ (x(x + 1)) ^{2} + {(y - 2)}^{2} = 0$
так как квадрат любого выражения ≥0, то сумма квадратов выражений равна нулю, когда каждое из выражений одновременно равно нулю
$\left \{ {{x = 0} \atop {y - 2= 0}} \right.$
и $\left \{ {{x + 1= 0} \atop {y - 2= 0}} \right.$
откуда получаем решения

$\\ \left \{ {{x_1= 0} \atop {y_1= 2}} \right. \left \{ {{x_2= - 1} \atop {y_2= 2}} \right.$

Answer 2

The equation you provided is a quadratic equation in two variables, x and y. It is in the form of a conic section, specifically an ellipse. However, there seems to be an issue with the equation you provided, as an ellipse equation should not equal 0. Typically, the equation for an ellipse is in the form:

$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$

Where (h, k) is the center of the ellipse, and 'a' and 'b' are the semi-major and semi-minor axes lengths, respectively.

If you meant to provide an equation for an ellipse, please double-check and correct the equation. If you have any other questions or need further assistance, please let me know.