Question

(3/(9-m^2 )+(2m-1)/(m-3)-(m^2-4)/(m^2+6m+9)×(m+3)/(m-2))÷m/(m-3)

Answer 1

$\tt\displaystyle (\frac{3}{9-m^2} +\frac{2m-1}{m-3}-\frac{m^2-4}{m^2+6m+9}\cdot \frac{m+3}{m-2}): \frac{m}{m-3}=\\\\ (\frac{-3}{m^2-9}+\frac{2m-1}{m-3}-\frac{(m-2)(m+2)(m+3)}{(m+3)^2(m-2)})\cdot \frac{m-3}{m}=\\\\ (\frac{-3+(2m-1)(m+3)}{(m-3)(m+3)}-\frac{m+2}{m+3})\cdot \frac{m-3}{m}=\\\\ (\frac{2m^2+6m-m-3-3-(m^2+2m-3m-6)}{(m+3)(m-3)})\cdot \frac{m-3}{m}=\\\\ \frac{(m^2+6m)(m-3)}{(m+3)(m-3)m}=\frac{m+6}{m+3}$

Answer 2

Before we can simplify the expression, we need to first find any potential values of m that would make any of the denominators equal to zero (since dividing by zero is undefined).

Starting with the first fraction:

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3/(9-m^2)

This denominator can be factored as the difference of two squares:

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3/((3+m)(3-m))

So we see that the denominator will be zero when m = 3 or m = -3.

Moving on to the second fraction:

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(2m-1)/(m-3)

This denominator will be zero when m = 3.

The third fraction is more complex:

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(m^2-4)/(m^2+6m+9)×(m+3)/(m-2)

The denominator m^2+6m+9 can be factored as a perfect square:

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(m+3)^2

So we can simplify the fraction:

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(m^2-4)/(m+3)^2 × (m+3)/(m-2)

Canceling out the (m+3) term from the numerator and denominator, we get:

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(m-2)(m+2)/(m+3)

So this fraction will be undefined when m = -3.

Finally, let's simplify the expression:

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(3/((3+m)(3-m)) + (2m-1)/(m-3) - (m-2)(m+2)/(m+3)) × m/(m-3)

We can start by combining the first and third fractions:

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(3(m-2)(m+2) - (3+m)(3-m)(m-2)(m+2) + (2m-1)(3+m)(3-m)) / ((3+m)(3-m)(m-3)(m+3))

Simplifying the numerator:

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(3m-6)(m+2) - (9-m^2)(m-2)(m+2) + (2m-1)(9-m^2)

Multiplying out the factors:

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3m^2 - 12m - 6 - (9-m^2)(m^2-4) + 18m^2 - 9m + (2m-1)(9-m^2)

Expanding further:

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3m^2 - 12m - 6 - (9m^4 - 18m^2 - 36) + 18m^2 - 9m + (18m^2 - 2m^3 - 9m^2 + 18m - 9)

Combining like terms:

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-9m^4 + 2m^3 + 21m^2 - 3m - 27

So the simplified expression is:

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(-9m^4 + 2m^3 + 21m^2 - 3m - 27)/(m^2-9)(m-3)