Quy đồng mẫu các phân thức sau:
a) \({{x - 1} \over {{x^2} - 9}}\) và \({{2xy + 1} \over {2x + 6}}\) ;
b) \({{3x + y} \over {{y^2} - 2xy + {x^2}}}\) và \({{y + 1} \over {2x - 2y}}\) .
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\(\eqalign{ & a)\,\,{x^2} - 9 = \left( {x - 3} \right)\left( {x + 3} \right) \cr & \,\,\,\,\,2x + 6 = 2\left( {x + 3} \right) \cr & MTC = 2\left( {x - 3} \right)\left( {x + 3} \right) \cr & {{x - 1} \over {{x^2} - 9}} = {{x - 1} \over {\left( {x - 3} \right)\left( {x + 3} \right)}} = {{2\left( {x - 1} \right)} \over {2\left( {x - 3} \right)\left( {x + 3} \right)}} \cr & {{2xy + 1} \over {2x + 6}} = {{2xy + 1} \over {2\left( {x + 3} \right)}} = {{\left( {2xy + 1} \right)\left( {x - 3} \right)} \over {2\left( {x - 3} \right)\left( {x + 3} \right)}} \cr & b)\,\,{y^2} - 2xy + {x^2} = {\left( {x - y} \right)^2} \cr & \,\,\,\,\,2x - 2y = 2\left( {x - y} \right) \cr & MTC = 2{\left( {x - y} \right)^2} \cr & {{3x + y} \over {{y^2} - 2xy + {x^2}}} = {{3x + y} \over {{{\left( {x - y} \right)}^2}}} = {{2\left( {3x + y} \right)} \over {2{{\left( {x - y} \right)}^2}}} \cr & {{y + 1} \over {2x - 2y}} = {{y + 1} \over {2\left( {x - y} \right)}} = {{\left( {y + 1} \right)\left( {x - y} \right)} \over {2{{\left( {x - y} \right)}^2}}} \cr} \)