Question

(x²-2x)²-3(x²-2x)-4=0

Answer 1

Решение

(x²-2x)²-3(x²-2x)-4=0

Используем подстановку t=x²-2x

t²-3t-4=0

t=4

t=-1

x²-2x=4

x²-2x=-1

x=1+√5

x=1-√5

x=1

Ответ

х1=1-√5 , х2=1 , х3=1+√5

Answer 2

Это можно найти в Photomath,скачайте)

Answer 3

Let's solve the quadratic equation:

(x²-2x)² - 3(x²-2x) - 4 = 0

To make this equation more manageable, let's introduce a substitution. Let's set y = x²-2x. Then the equation becomes:

y² - 3y - 4 = 0

Now, we can solve this quadratic equation for y. The equation can be factored:

(y - 4)(y + 1) = 0

Now, set each factor to zero and solve for y:

1. y - 4 = 0 y = 4

2. y + 1 = 0 y = -1

We have two possible values for y: y = 4 and y = -1.

Now, recall that y = x²-2x. So, we can set each y-value equal to x²-2x and solve for x:

1. x²-2x = 4

To solve this quadratic equation, we'll move everything to one side:

x² - 2x - 4 = 0

Now, we can either use the quadratic formula or complete the square to find the solutions for x. Let's use the quadratic formula:

x = (-(-2) ± √((-2)² - 4 * 1 * (-4))) / (2 * 1) x = (2 ± √(4 + 16)) / 2 x = (2 ± √20) / 2 x = (2 ± 2√5) / 2

Now, we can simplify the expression:

x = (1 ± √5)

So, we have two solutions for x: x = 1 + √5 and x = 1 - √5.

2. x²-2x = -1

Again, let's move everything to one side:

x² - 2x + 1 = 0

Now, we can recognize that this is a perfect square:

(x - 1)² = 0

Taking the square root of both sides:

x - 1 = 0

Now, solve for x:

x = 1

So, we have another solution for x: x = 1.

In conclusion, the solutions to the original equation are: x = 1 + √5 x = 1 - √5 x = 1