(√2-1)^х+(√2+1)^х-2=0

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This expression can be written as a binomial expansion. Let's let y = (√2 - 1)^x. Then, we have:

(√2 - 1)^x = y

(√2 + 1)^x = 1/y

Substitute these into the equation:

y + 1/y - 2 = 0

Multiplying through by y, we get:

y^2 + 1 - 2y = 0

Rearranging, we get a quadratic equation:

y^2 - 2y + 1 = 0

This equation factors as:

(y - 1)^2 = 0

So, y = 1

Now, substitute back to get the value of x:

(√2 - 1)^x = 1

Since 1 to any power is still 1, we have:

(√2 - 1)^x = 1

So, x = 0

Therefore, the solution to the equation (√2 - 1)^x + (√2 + 1)^x - 2 = 0 is x = 0.