Решите пж уравнения 1.(x^2-4x)^2+9(x^-4x)=-202. y^4-9^2+20=0

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Expand the terms:

(x^4 - 8x^3 + 16x^2) + 9x - 36x = -202
x^4 - 8x^3 + 16x^2 + 9x - 36x + 202 = 0
x^4 - 8x^3 + 16x^2 - 27x + 202 = 0

Unfortunately, this equation does not appear to have a simple solution and would require the use of numerical methods to solve.

This is a quadratic equation in terms of y^2. Let's substitute z = y^2:

z^2 - 9z + 20 = 0
(z - 4)(z - 5) = 0

Now, solve for z:

z - 4 = 0 or z - 5 = 0
z = 4 or z = 5

Now, substitute back y^2 for z:

y^2 = 4 or y^2 = 5

Taking the square root of both sides:

y = ±2 or y = ±√5

Therefore, the solutions for the second equation are y = 2, y = -2, y = √5, and y = -√5.