a) \({{x – 1} \over 2}\) và \({{{x^2}} \over {{x^2} – 16}}\) ;
b) \({{x + y} \over {{y^3} – 3{x^2}y + 3x{y^2} – {x^3}}}\) và \({1 \over {{x^2} – xy}}\)
Advertisements (Quảng cáo)
\(\eqalign{ & a)\,\,{x^2} – 16 = \left( {x – 4} \right)\left( {x + 4} \right) \cr & MTC = 2\left( {x – 4} \right)\left( {x + 4} \right) \cr & {{x – 1} \over 2} = {{\left( {x – 1} \right)\left( {x – 4} \right)\left( {x + 4} \right)} \over {2\left( {x – 4} \right)\left( {x + 4} \right)}} \cr & {{{x^2}} \over {{x^2} – 16}} = {{{x^2}} \over {\left( {x – 4} \right)\left( {x + 4} \right)}} = {{2{x^2}} \over {2\left( {x – 4} \right)\left( {x + 4} \right)}} \cr & b)\,\,{y^3} – 3x{y^2} + 3{x^2}y – {x^3} = {\left( {y – x} \right)^3} \cr & \,\,\,\,\,{x^2} – xy = – x\left( {y – x} \right) \cr & MTC = x{\left( {y – x} \right)^3} \cr & {{x + y} \over {{y^3} – 3x{y^2} + 3{x^2}y – {x^3}}} = {{x + y} \over {{{\left( {y – x} \right)}^3}}} = {{x\left( {x + y} \right)} \over {x{{\left( {y – x} \right)}^3}}} \cr & {1 \over {{x^2} – xy}} = {1 \over { – x\left( {y – x} \right)}} = {{\left( { – 1} \right){{\left( {y – x} \right)}^2}} \over {x{{\left( {y – x} \right)}^3}}} = {{ – {{\left( {y – x} \right)}^2}} \over {x{{\left( {y – x} \right)}^3}}} \cr} \)