Ta có: \(\displaystyle A = {5 \over {2m + 1}}\) và \(\displaystyle B = {4 \over {2m - 1}}\) ĐKXĐ: \(\displaystyle m \ne \pm {1 \over 2}\)
a)
\(\displaystyle \eqalign{ & 2A + 3B = 0 \cr & \Leftrightarrow 2.{5 \over {2m + 1}} + 3.{4 \over {2m - 1}} = 0 \cr & \Leftrightarrow {{10} \over {2m + 1}} +{{12} \over {2m - 1}} = 0 \cr} \)
\(\displaystyle \Leftrightarrow {{10\left( {2m - 1} \right)} \over {\left( {2m + 1} \right)\left( {2m - 1} \right)}} \) \(\displaystyle+ {{12\left( {2m + 1} \right)} \over {\left( {2m + 1} \right)\left( {2m - 1} \right)}} = 0 \)
\(\displaystyle \eqalign{ & \Rightarrow 10\left( {2m - 1} \right) + 12\left( {2m + 1} \right) = 0 \cr & \Leftrightarrow 20m - 10 + 24m + 12 = 0 \cr & \Leftrightarrow 44m + 2 = 0 \cr} \)
\(\displaystyle \Leftrightarrow m = - {1 \over {22}}\) (thỏa mãn)
Vậy \(\displaystyle m = - {1 \over {22}}\) thì \(2A + 3B = 0.\)
b) \(\displaystyle.B = A + {\rm B} \)
\(\displaystyle \Leftrightarrow {5 \over {2m + 1}}.{4 \over {2m - 1}} = {5 \over {2m + 1}} \) \(\displaystyle + {4 \over {2m - 1}} \)
\(\displaystyle \Leftrightarrow {{20} \over {\left( {2m + 1} \right)\left( {2m - 1} \right)}}\) \(\displaystyle= {{5\left( {2m - 1} \right)} \over {\left( {2m + 1} \right)\left( {2m - 1} \right)}} \) \(\displaystyle + {{4\left( {2m + 1} \right)} \over {\left( {2m + 1} \right)\left( {2m - 1} \right)}} \)
\(\displaystyle\eqalign{ & \Rightarrow 20 = 5\left( {2m - 1} \right) + 4\left( {2m + 1} \right) \cr & \Leftrightarrow 20 = 10m - 5 + 8m + 4 \cr & \Leftrightarrow 18m = 21 \cr} \)
\(\displaystyle \;\;\Leftrightarrow m = {7 \over 6}\) (thỏa mãn)
Vậy \(\displaystyle m = {7 \over 6}\) thì \(A.B = A + B.\)
