Thực hiện phép chia phân thức :
a. \({{{x^2} - 5x + 6} \over {{x^2} + 7x + 12}}:{{{x^2} - 4x + 4} \over {{x^2} + 3x}}\)
b. \({{{x^2} + 2x - 3} \over {{x^2} + 3x - 10}}:{{{x^2} + 7x + 12} \over {{x^2} - 9x + 14}}\)
a. \({{{x^2} - 5x + 6} \over {{x^2} + 7x + 12}}:{{{x^2} - 4x + 4} \over {{x^2} + 3x}}\)\( = {{{x^2} - 5x + 6} \over {{x^2} + 7x + 12}}.{{{x^2} + 3x} \over {{x^2} - 4x + 4}}\)
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\( = {{\left( {{x^2} - 5x + 6} \right).x\left( {x + 3} \right)} \over {\left( {{x^2} + 7x + 12} \right){{\left( {x - 2} \right)}^2}}} = {{\left( {{x^2} - 2x - 3x + 6} \right).x\left( {x + 3} \right)} \over {\left( {{x^2} + 3x + 4x + 12} \right){{\left( {x - 2} \right)}^2}}}\)
\( = {{\left[ {x\left( {x - 2} \right) - 3\left( {x - 2} \right)} \right].x\left( {x + 3} \right)} \over {\left[ {x\left( {x + 3} \right) + 4\left( {x + 3} \right)} \right]{{\left( {x - 2} \right)}^2}}}\)
\( = {{x\left( {x - 2} \right)\left( {x - 3} \right)\left( {x + 3} \right)} \over {\left( {x + 3} \right)\left( {x + 4} \right){{\left( {x - 2} \right)}^2}}} = {{x\left( {x - 3} \right)} \over {\left( {x + 4} \right)\left( {x - 2} \right)}}\)
b. \({{{x^2} + 2x - 3} \over {{x^2} + 3x - 10}}:{{{x^2} + 7x + 12} \over {{x^2} - 9x + 14}}\)\( = {{{x^2} + 2x - 3} \over {{x^2} + 3x - 10}}.{{{x^2} - 9x + 14} \over {{x^2} + 7x + 12}}\)
\(\eqalign{ & = {{\left( {{x^2} + 2x - 3} \right)\left( {{x^2} - 9x + 14} \right)} \over {\left( {{x^2} + 3x - 10} \right)\left( {{x^2} + 7x + 12} \right)}} = {{\left( {{x^2} + 3x - x - 3} \right)\left( {{x^2} - 7x - 2x + 14} \right)} \over {\left( {{x^2} + 5x - 2x + 10} \right)\left( {{x^2} + 3x + 4x + 12} \right)}} \cr & = {{\left[ {x\left( {x + 3} \right) - \left( {x + 3} \right)} \right]\left[ {x\left( {x - 7} \right) - 2\left( {x - 7} \right)} \right]} \over {\left[ {x\left( {x + 5} \right) - 2\left( {x + 5} \right)} \right]\left[ {x\left( {x + 3} \right) + 4\left( {x + 3} \right)} \right]}} \cr & = {{\left( {x + 3} \right)\left( {x - 1} \right)\left( {x - 7} \right)\left( {x - 2} \right)} \over {\left( {x + 5} \right)\left( {x - 2} \right)\left( {x + 3} \right)\left( {x + 4} \right)}} = {{\left( {x - 1} \right)\left( {x - 7} \right)} \over {\left( {x + 5} \right)\left( {x + 4} \right)}} \cr} \)