Question

( √ x − 2 − 3 ) ∗ ( 3 x 2 − 11 x + 30 − 1 ) ∗ l n ( x + 9 ) = 0 .

Answer 1

Відповідь:

=2

Покрокове пояснення:

я не придумав

Answer 2

To solve the equation (√x - 2 - 3) * (3x^2 - 11x + 30 - 1) * ln(x + 9) = 0, we need to find the values of x that satisfy this equation.

To do that, we can break down the equation and solve each part separately:

1. (√x - 2 - 3) = 0 Simplifying the equation, we get: √x - 5 = 0

   Squaring both sides of the equation, we have: (√x)^2 - 10(√x) + 25 = 0 x - 10√x + 25 = 0

   Let's substitute a variable to simplify the equation: Let u = √x u^2 - 10u + 25 = 0

   Factoring the equation, we get: (u - 5)(u - 5) = 0

   Since (u - 5)(u - 5) = 0, we have two cases: u - 5 = 0, which gives u = 5 u - 5 = 0, which gives u = 5

   Substituting back u = √x, we get: √x = 5

   Squaring both sides again: (√x)^2 = 5^2 x = 25

   So, the first solution is x = 25.

2. (3x^2 - 11x + 30 - 1) = 0 Simplifying the equation, we have: 3x^2 - 11x + 29 = 0

   Unfortunately, this quadratic equation does not factor easily. We can use the quadratic formula to find the solutions:

   x = (-b ± √(b^2 - 4ac)) / (2a)

   For this equation, a = 3, b = -11, and c = 29. Substituting these values into the quadratic formula:

   x = (-(-11) ± √((-11)^2 - 4(3)(29))) / (2(3)) x = (11 ± √(121 - 348)) / 6 x = (11 ± √(-227)) / 6

   The term inside the square root, -227, is negative, which means there are no real solutions for this part of the equation.

   Therefore, the second part of the equation has no real solutions.

3. ln(x + 9) = 0 To solve this equation, we need to find the value of x that makes the natural logarithm of (x + 9) equal to zero.

   ln(x + 9) = 0 x + 9 = e^0 x + 9 = 1 x = 1 - 9 x = -8

   So, the third solution is x = -8.

Therefore, the solutions to the equation (√x - 2 - 3) * (3x^2 - 11x + 30 - 1) * ln(x + 9) = 0 are x = 25 and x = -8.