1)(2x+5)^2=16+(x-3)^22)(x-1)^2=17-(3x+4)^2

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вопрос

1) Expanding the left side of the equation:
(2x+5)^2 = (2x+5)(2x+5) = 4x^2 + 10x + 10x + 25
= 4x^2 + 20x + 25

Expanding the right side of the equation:
16 + (x-3)^2 = 16 + (x^2 - 6x + 9)
= 16 + x^2 - 6x + 9
= x^2 - 6x + 25

Setting the left and right sides equal to each other:
4x^2 + 20x + 25 = x^2 - 6x + 25
Combine like terms:
3x^2 + 26x = 0

Now we need to solve for x. Factoring out an x from the equation:
x(3x + 26) = 0
Setting each factor equal to zero:
x = 0 or 3x + 26 = 0
If we solve for x in the second equation:
3x = -26
x = -26/3

Therefore, the solutions are x = 0 and x = -26/3.

2) Expanding the left side of the equation:
(x-1)^2 = (x-1)(x-1) = x^2 - x - x + 1
= x^2 - 2x + 1

Expanding the right side of the equation:
17 - (3x+4)^2 = 17 - (9x^2 + 24x + 16)
= 17 - 9x^2 - 24x - 16
= -9x^2 - 24x + 1

Setting the left and right sides equal to each other:
x^2 - 2x + 1 = -9x^2 - 24x + 1

Combining like terms:
10x^2 + 22x = 0

Factoring out an x from the equation:
x(10x + 22) = 0
Setting each factor equal to zero:
x = 0 or 10x + 22 = 0
If we solve for x in the second equation:
10x = -22
x = -22/10
x = -11/5

The solutions to the equation are x = 0 and x = -11/5.