Question

1\(a-2) + 2\(a2+1)=5\(a3-2a2+a-2)

Answer 1

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Answer 2

Let's simplify and solve the given equation:

1/(a-2) + 2/(a^2+1) = 5/(a^3-2a^2+a-2)

First, let's get rid of the fractions by multiplying both sides of the equation by the least common multiple of the denominators, which is (a-2)(a^2+1)(a^3-2a^2+a-2):

(a-2)(a^2+1)(a^3-2a^2+a-2) * 1/(a-2) + (a-2)(a^2+1)(a^3-2a^2+a-2) * 2/(a^2+1) = (a-2)(a^2+1)(a^3-2a^2+a-2) * 5/(a^3-2a^2+a-2)

Simplifying this expression, we get:

(a^2+1)(a^3-2a^2+a-2) + 2(a-2)(a^3-2a^2+a-2) = 5(a-2)(a^2+1)

Expanding and simplifying the left-hand side of the equation, we get:

a^5 - a^4 - 3a^3 + 3a^2 + 6a - 4 + 2a^4 - 8a^3 + 10a^2 - 8a + 4 = 5a^3 - 10a^2 + 5a - 10

Combining like terms, we get:

a^5 + a^4 - a^3 - 11a^2 + 3a - 10 = 0

Now, we can try to factor the left-hand side of the equation:

(a^2 + 1)(a^3 - 10a^2 + 3a - 10) = 0

So either a^2 + 1 = 0 or a^3 - 10a^2 + 3a - 10 = 0.

The equation a^2 + 1 = 0 has no real solutions, so we can discard it.

To solve the equation a^3 - 10a^2 + 3a - 10 = 0, we can use numerical methods or approximate solutions. One possible method is to use Newton's method, starting with an initial guess, for example, a = 2:

a1 = a0 - f(a0)/f'(a0) = 2 - (-16)/(-14) = 30/14

a2 = a1 - f(a1)/f'(a1) = 30/14 - (-6320/3436)/(153/98) = 406/153

a3 = a2 - f(a2)/f'(a2) = 406/153 - (-247388/107266)/(2247/1378) = 2458/1073

a4 = a3 - f(a3)/f'(a3) = 2458/1073 - (-12447008/6896789)/(10954/6661) = 6444/2731

a5 = a4 - f(a4)/f'(a4) = 6444/2731 - (-8827469408/5996142723)/(25192/15271) = 1.6972

So one solution of the equation is approximately a = 1.6972. We can check that this value satisfies the original equation by substituting it: