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Simplify $\; \large\frac{6m-3}{2mn} \;\div \; \large\frac{4m^{2}-1}{8m^{2}}$

$(a)\;\large\frac{6m}{2m+1}\qquad(b)\;\large\frac{12m}{n(2m+1)}\qquad(c)\;\large\frac{12m}{2m+1}\qquad(d)\;\large\frac{6m}{n(2m+1)}$

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Answer : $\;\large\frac{12m}{2m+1}$
$\; \large\frac{6m-3}{2mn} \;\div \; \large\frac{4m^{2}-1}{8m^{2}} = \large\frac{12m}{2m+1}$
answered Apr 12, 2014 by yamini.v
 

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