Question

(2х-1)^4+(2х-1)^2-2=0

Answer 1

Ответ:

х = 1

Объяснение:

(2х-1)⁴+(2х-1)²-2 = 0

Пусть (2х-1)² = y, y≥0, тогда

у²+у-2 = 0

D = 1²-4*(-2) = 1+8 = 9 = 3²

y1 = (-1+3)/(2*1) = 2/2 = 1

y2 = (-1-3)/(2*1) = -4/2 = -2 - не подходит по условию

Вернёмся к замене

Если у = 1,то

2х-1 = 1

2х = 1+1

2х = 2

х = 1

Answer 2

To solve the equation (2x-1)^4 + (2x-1)^2 - 2 = 0, we can use a substitution to simplify the expression. Let's substitute (2x-1) with another variable, say "y".

Let y = (2x-1).

Now, we can rewrite the equation in terms of y:

y^4 + y^2 - 2 = 0.

This is now a quadratic equation in terms of y. To solve it, we can use factoring or the quadratic formula. Let's use factoring:

(y^2 + 2)(y^2 - 1) = 0.

Now, we can solve each factor separately:

1. y^2 + 2 = 0: y^2 = -2 This equation has no real solutions since the square of any real number is non-negative.

2. y^2 - 1 = 0: y^2 = 1 Taking the square root of both sides, we have: y = ±1

Now, substitute back y = (2x-1):

For y = 1: 2x - 1 = 1 2x = 2 x = 1

For y = -1: 2x - 1 = -1 2x = 0 x = 0

Therefore, the solutions to the equation (2x-1)^4 + (2x-1)^2 - 2 = 0 are x = 1 and x = 0.