Quy đồng mẫu thức các phân thức sau:
a) \({5 \over {2x + 2}}\) và \({9 \over {{x^2} – 1}}\) ;
b) \({1 \over {4 – 2x}}\) và \({3 \over {{x^2} – 4}}\) .
\(\eqalign{ & a)\,\,2x + 2 = 2\left( {x + 1} \right); \cr & \,\,\,\,\,\,\,{x^2} – 1 = \left( {x – 1} \right)\left( {x + 1} \right) \cr & MTC = 2\left( {x + 1} \right)\left( {x – 1} \right) \cr & {5 \over {2x + 2}} = {5 \over {2\left( {x + 1} \right)}} = {{5\left( {x – 1} \right)} \over {2\left( {x + 1} \right)\left( {x – 1} \right)}} \cr & {9 \over {{x^2} – 1}} = {9 \over {\left( {x – 1} \right)\left( {x + 1} \right)}} = {{9.2} \over {2\left( {x + 1} \right)\left( {x – 1} \right)}} = {{18} \over {2\left( {x + 1} \right)\left( {x – 1} \right)}} \cr & b)\,\,4 – 2x = – 2\left( {x – 2} \right) \cr & \,\,\,\,\,\,\,{x^2} – 4 = \left( {x – 2} \right)\left( {x + 2} \right) \cr & MTC = 2\left( {x – 2} \right)\left( {x + 2} \right) \cr & {1 \over {4 – 2x}} = {1 \over { – 2\left( {x – 2} \right)}} = {{ – 1\left( {x + 2} \right)} \over {2\left( {x – 2} \right)\left( {x + 2} \right)}} = {{ – x – 2} \over {2\left( {x – 2} \right)\left( {x + 2} \right)}} \cr & {3 \over {{x^2} – 4}} = {3 \over {\left( {x – 2} \right)\left( {x + 2} \right)}} = {{3.2} \over {2\left( {x – 2} \right)\left( {x + 2} \right)}} = {6 \over {2\left( {x – 2} \right)\left( {x + 2} \right)}} \cr} \)