Question

(x-5)⁴+(x-5)²-20=о ответ с решением срочно

Answer 1

Пусть (x-5)^2 = t, тогда t^2 + t - 20 =0
D= 1-4*1*(-20)=81=9^2
t1=(-1-9)\2=-5, (x-5)^2=-5 не имеет смысла

t2=(-1+9)\2=4, (x-5)^2=4
  x^2 - 10x + 21 = 0
  D=100-4*1*21=16=4^2
  x1=(10-4)\2=3
  x2=(10+4)\2=7
Ответ: 3 и 7.

Answer 2

Solving the Equation (x-5)⁴ + (x-5)² - 20 = 0

To solve the equation (x-5)⁴ + (x-5)² - 20 = 0, we can use a substitution to simplify the equation and then solve for the variable.

Let's start by making a substitution: Let y = (x-5)²

Now, the equation becomes: y² + y - 20 = 0

We can solve this quadratic equation for y using the quadratic formula: y = (-b ± √(b² - 4ac)) / (2a)

Where a = 1, b = 1, and c = -20.

Plugging these values into the quadratic formula, we get: y = (-1 ± √(1² - 4*1*(-20))) / (2*1) y = (-1 ± √(1 + 80)) / 2 y = (-1 ± √81) / 2 y = (-1 ± 9) / 2

So, we have two possible values for y: 1. y = (-1 + 9) / 2 = 8 / 2 = 4 2. y = (-1 - 9) / 2 = -10 / 2 = -5

Now that we have the values for y, we can substitute back to find the values for x: For y = 4: (x-5)² = 4 Taking the square root of both sides, we get: x-5 = ±2 So, we have two possible equations: 1. x-5 = 2, which gives x = 7 2. x-5 = -2, which gives x = 3

For y = -5: (x-5)² = -5 This equation has no real solutions since the square of a real number is always non-negative.

Therefore, the solutions to the original equation (x-5)⁴ + (x-5)² - 20 = 0 are: x = 7 and x = 3.