Решите уравнения 1)12x^2-25=0 2)(x-1)^2+3 3)√16x^2-24x+9=11

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вопрос

1) 12x^2 - 25 = 0
12x^2 = 25
x^2 = 25/12
x = ±√(25/12)
x = ±(5/√12)
x = ±(5√3)/6

2) (x - 1)^2 + 3 = 0
(x - 1)^2 = -3
There are no real solutions for this equation because a square of a real number cannot be negative.

3) √16x^2 - 24x + 9 = 11
4x - 24x + 9 = 11
4x^2 - 24x - 2 = 0
2(2x^2 - 12x - 1) = 0
Using the quadratic formula, we find:
x = (12 ± √(12^2 + 8))/4
x = (12 ± √(144 + 8))/4
x = (12 ± √152)/4
x = (12 ± 2√38)/4
x = 3 ± √38/2

So, the solutions are x = 3 + √38/2 or x = 3 - √38/2.