1) |x+3|+|y-2|+|3^2^x-27|=0 2) x^2^+y^2^+34=6x+10y

Предыдущий
вопрос Следующий
вопрос

1) The first equation has a solution when each absolute value term equals zero.

Given |x+3| = 0, we find that x = -3.
Given |y-2| = 0, we find that y = 2.
Given |3^2^x-27| = 0, we find that 3^2^x = 27.
Solving for x in 3^2^x = 27, we see that x = 1.

Thus, the solution to the first equation is x = -3, y = 2, x = 1.

2) The second equation is a circle formula. We can rewrite it as:
x^2 - 6x + y^2 - 10y + 34 = 0

Rearranging terms, we see:
(x^2 - 6x + 9) + (y^2 - 10y + 25) = - 34 + 9 + 25
(x - 3)^2 + (y - 5)^2 = 0

This is the equation of a circle centered at (3, 5) with radius 0. Therefore, the solution to the second equation is simply the point (3, 5).