Quy đồng mẫu thức các phân thức sau:
a) \({1 \over {3x – 9}}\) và \({2 \over {{x^2} – 6x + 9}}\) ;
b) \({7 \over {4 – 2x}}\) và \({2 \over {{x^2} – 4x + 4}}\) .
\(\eqalign{ & a)\,\,3x – 9 = 3\left( {x – 3} \right) \cr & \,\,\,\,\,\,{x^2} – 6x + 9 = {\left( {x – 3} \right)^2} \cr & MTC:\,\,3{\left( {x – 3} \right)^2} \cr & {1 \over {3x – 9}} = {1 \over {3\left( {x – 3} \right)}} = {{x – 3} \over {3{{\left( {x – 3} \right)}^2}}} \cr & {2 \over {{x^2} – 6x + 9}} = {2 \over {{{\left( {x – 3} \right)}^2}}} = {{2.3} \over {3{{\left( {x – 3} \right)}^2}}} = {6 \over {3{{\left( {x – 3} \right)}^2}}} \cr & b)\,\,4 – 2x = – 2\left( {x – 2} \right) \cr & \,\,\,\,\,\,{x^2} – 4x + 4 = {\left( {x – 2} \right)^2} \cr & MTC:\,\,2{\left( {x – 2} \right)^2} \cr & {7 \over {4 – 2x}} = {7 \over { – 2\left( {x – 2} \right)}} = {{ – 7\left( {x – 2} \right)} \over {2{{\left( {x – 2} \right)}^2}}} \cr & {2 \over {{x^2} – 4x + 4}} = {2 \over {{{\left( {x – 2} \right)}^2}}} = {{2.2} \over {2{{\left( {x – 2} \right)}^2}}} = {4 \over {2{{\left( {x – 2} \right)}^2}}} \cr} \)