Question

(х-4)^4-2(x^2-8x+16)-8=0

Answer 1

(х-4)^4-2(х-4) ^2-8=0;
Замена (х-4) ^2 = у;
у ^2-2у-8=0;

у=4
у=-2

Обратная замена

х=2
х=-2
х= корень 2
х= - корень 2

Если не сложно, можно лучший ответ?

Answer 2

(X-4)^2 можно взять как t , чтоб было легче

Answer 3

To solve the equation (х-4)^4-2(x^2-8x+16)-8=0, we can use the substitution u = x-4 to simplify the expression and then solve for u.

Substituting u = x-4, we have:

(u)^4 - 2(u^2 - 8u + 16) - 8 = 0

Simplifying the expression:

u^4 - 2u^2 + 16u^2 - 32u + 8 - 8 = 0

u^4 + 14u^2 - 32u = 0

Factorizing:

u(u^3 + 14u - 32) = 0

Solving for u, we get:

u = 0 or u^3 + 14u - 32 = 0

If u = 0, then x - 4 = 0, which implies x = 4.

To solve the cubic equation, we can use the Rational Root Theorem to find possible rational roots, and then use synthetic division to test each root until we find a root that makes the equation equal to zero.

The possible rational roots of u^3 + 14u - 32 = 0 are ±1, ±2, ±4, ±8, ±16, ±32. Testing these roots using synthetic division, we find that u = 2 is a root. Therefore, we can factorize the cubic equation as:

u^3 + 14u - 32 = (u - 2)(u^2 + 2u + 16)

Using the quadratic formula to solve for the roots of the quadratic factor, we get:

u = -1 ± 3i

Substituting back u = x-4, we get:

x = 4, 2 + 4i, 2 - 4i, -1 + 4i, -1 - 4i

Therefore, the solutions to the equation (х-4)^4-2(x^2-8x+16)-8=0 are x = 4, 2 + 4i, 2 - 4i, -1 + 4i, -1 - 4i.