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3x+1/x^(2)-1=2/x-2+x/x-1

Sagot :

mathstudent55

Answer:

x = 0 and 3 other real roots. See below.

Step-by-step explanation:

This is what you wrote:

[tex] 3x + \dfrac{1}{x^2} - 1 = \dfrac{2}{x} - 2 + \dfrac{x}{x} - 1 [/tex]

I don't think that is what you meant to write.

I think you meant to write this:

[tex] 3x + \dfrac{1}{x^2 - 1} = \dfrac{2}{x - 2} + \dfrac{x}{x - 1} [/tex]

I'm going to answer the second equation because that is what I think you meant.

[tex] 3x + \dfrac{1}{(x - 1)(x + 1)} = \dfrac{2}{x - 2} + \dfrac{x}{x - 1} [/tex]

[tex] 3x(x - 1)(x + 1)(x - 2) + (x - 2) = 2(x - 1)(x + 1) + x(x + 1)(x - 2) [/tex]

[tex] (3x^2 - 6x)(x^2 - 1) + x - 2 = 2x^2 - 2 + x(x^2 - x - 2) [/tex]

[tex] 3x^4 - 3x^2 - 6x^3 + 6x + x - 2 = 2x^2 - 2 + x^3 - x^2 - 2x [/tex]

[tex] 3x^4 - 7x^3 - 4x^2 + 9x = 0 [/tex]

[tex] x(3x^3 - 7x^2 - 4x + 9) = 0 [/tex]

[tex] f(x) = 3x^3 - 7x^2 - 4x + 9 [/tex]

[tex] f(-3) = -123 [/tex]

[tex] f(-2) = -35 [/tex]

[tex] f(-1) = 3 [/tex]

[tex] f(0) = 9 [/tex]

[tex] f(1) = 1 [/tex]

[tex] f(2) = -3 [/tex]

[tex] f(3) = 15 [/tex]

One root is x = 0.

There is a root between x = -2 and x = -1.

There is a root between x = 1 and x = 2.

There is a root between x = 2 and x = 3.

Plot the graph of f(x) = 3x^3 - 7x^2 - 4x + 9 and try to read the other three roots.

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