Question

Помогите довести что при все значениях исполняеться функция: х в квадрате +у в квадрате+ 8х-10у+42>0

Answer 1

Ну смотри. Доказывать можно как графически, так и алгебраически. Просто я тебе не смогу график нарисовать, а алгебраически сейчас распишу. 
Тут просто можно сделать таким образом:
$x^2+y^2+8x-10y+42=(x^2+8x+16)+(y^2-10y+25)+1$
Тут просто выделяем полный квадрат и получаем:
$(x+4)^2 + (y-5)^2+1$
Т.к. квадраты всегда больше или равны нулю, то и эта сума будет больше нуля:
$(x+4)^2\ \textgreater \ 0; (y-5)^2\ \textgreater \ 0; 1\ \textgreater \ 0$
$(x+4)^2+(y-5)^2+1 \ \textgreater \ 0$

Answer 2

Checking the Inequality for All Values of x and y

To determine the values for which the function \( f(x, y) = x^2 + y^2 - 8x - 10y + 42 > 0 \), we can analyze the inequality and find its solution set.

Analyzing the Inequality

The inequality \( x^2 + y^2 - 8x - 10y + 42 > 0 \) represents a region in the xy-plane where the function \( f(x, y) \) is positive. To find the values of x and y for which this inequality holds true, we can consider the equation of a circle and use techniques from analytic geometry.

Completing the Square

To simplify the inequality and express it in a more manageable form, we can complete the square for both x and y terms. This will help us identify the center and radius of the circle represented by the inequality.

Completing the square for x: - \( x^2 - 8x = (x - 4)^2 - 16 \)

Completing the square for y: - \( y^2 - 10y = (y - 5)^2 - 25 \)

Rewriting the Inequality

By substituting the completed square expressions back into the original inequality, we get: \[ (x - 4)^2 + (y - 5)^2 - 16 - 25 + 42 > 0 \]

Simplifying, we have: \[ (x - 4)^2 + (y - 5)^2 + 1 > 0 \]

Conclusion

The inequality \( x^2 + y^2 - 8x - 10y + 42 > 0 \) simplifies to \( (x - 4)^2 + (y - 5)^2 + 1 > 0 \). Since the sum of squares is always non-negative, the inequality \( (x - 4)^2 + (y - 5)^2 + 1 > 0 \) holds true for all real values of x and y.

Therefore, the function \( f(x, y) = x^2 + y^2 - 8x - 10y + 42 \) is always greater than 0 for all values of x and y.

I hope this helps! If you have any further questions or need additional assistance, feel free to ask.