(2) \( \left(\frac{2}{m^{2}-m}-\frac{2 m}{1-m^{2}}\right) \cdot \frac{2 m^{2}+2 m}{m^{3}-1}+\frac{4}{m-1}= \)
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Beyond the Answer
To simplify the given expression \( \left(\frac{2}{m^{2}-m}-\frac{2 m}{1-m^{2}}\right) \cdot \frac{2 m^{2}+2 m}{m^{3}-1}+\frac{4}{m-1} \), we can follow these steps: 1. **Factor the components**: - Notice that \( m^2 - m = m(m-1) \). - The term \( 1 - m^2 = -(m^2 - 1) = -(m-1)(m+1) \). - The term \( m^3 - 1 = (m-1)(m^2 + m + 1) \). 2. **Rewrite the expression**: - The first term becomes: \[ \frac{2}{m(m-1)} - \frac{2m}{-(m-1)(m+1)} = \frac{2}{m(m-1)} + \frac{2m}{(m-1)(m+1)}. \] - Combine using a common denominator: - The common denominator is \( m(m-1)(m+1) \). - Both fractions can be expressed with this common denominator. 3. **Compute the combined first term**: \[ \frac{2(m+1) + 2m^2}{m(m-1)(m+1)} = \frac{2m^2 + 2m + 2}{m(m-1)(m+1)} = \frac{2(m^2 + m + 1)}{m(m-1)(m+1)}. \] 4. **Multiply by the second term**: - We then multiply by \( \frac{2m^2 + 2m}{m^3 - 1} = \frac{2m(m+1)}{(m-1)(m^2 + m + 1)} \). - So we have: \[ \left( \frac{2(m^2 + m + 1)}{m(m-1)(m+1)} \right) \cdot \left( \frac{2m(m+1)}{(m-1)(m^2 + m + 1)} \right). \] 5. **Simplifying the expression**: - Cancel \( m^2 + m + 1 \) from numerator and denominator: \[ \frac{4m(m+1)}{m(m-1)(m-1)(m+1)} = \frac{4}{(m-1)^2}. \] - Now we have: \[ \frac{4}{(m-1)^2} + \frac{4}{m-1}. \] 6. **Combine** the two fractions: - The common denominator is \( (m-1)^2 \): \[ \frac{4 + 4(m-1)}{(m-1)^2} = \frac{4 + 4m - 4}{(m-1)^2} = \frac{4m}{(m-1)^2}. \] 7. **Final result**: \[ \frac{4m}{(m-1)^2}. \] The final simplified result is: \[ \frac{4m}{(m-1)^2} . \]
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