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Old November 9th, 2008, 03:16 AM
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2) \displaystyle\frac{3a+1}{3a^2+7a+2}-\frac{a+2}{2a^2-a-10}=\frac{3a+1}{(3a+1)(a+2)}-\frac{a+2}{(a+2)(2a-5)}=

=\frac{1}{a+2}-\frac{1}{2a-5}=\frac{a-7}{(a+2)(2a-5)}

3) \displaystyle\frac{\frac{m^2-5m-24}{m^2-9)}}{\frac{m^2-10m+16}{m-3}}=\frac{m^2-5m-24}{m^2-9}\cdot\frac{m-3}{m^2-10m+16}=

=\frac{(m+3)(m-8)}{(m-3)(m+3)}\cdot\frac{m-3}{(m-2)(m-8)}=\frac{1}{m-2}
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