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Mkey [24]
3 years ago
14

Hi how yall doiiinnn

Mathematics
2 answers:
Maru [420]3 years ago
7 0

Pretty good, How about you

lukranit [14]3 years ago
5 0

Answer: heyyyyy I’m going good

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\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}

The degree of the numerator (4) is larger than the degree of the denominator (3), so first you need to divide. (Added screenshot of long division procedure.)

\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}=x-\dfrac{4x^2-17x+10}{x(x^2-3)}

Now the second term can be decomposed into partial fractions.

\dfrac{4x^2-17x+10}{x(x^2-3)}=\dfrac{r_1}x+\dfrac{r_2x+r_3}{x^2-3}
\dfrac{4x^2-17x+10}{x(x^2-3)}=\dfrac{r_1(x^2-3)+x(r_2x+r_3)}{x(x^2-3)}
4x^2-17x+10=r_1(x^2-3)+x(r_2x+r_3)
4x^2-17x+10=(r_1+r_2)x^2+r_3x-3r_1
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\implies\dfrac{4x^2-17x+10}{x(x^2-3)}=-\dfrac{10}{3x}+\dfrac{22x-51}{x^2-3}

So

\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}=x+\dfrac{10}{3x}-\dfrac{22x-51}{x^2-3}

3 0
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