Đề bài
Tìm mẫu thức chung rồi thực hiện phép tính:
a) \({4 \over {2 + y}} + {2 \over {y - 2}} + {{5y - 6} \over {4 - {y^2}}}\) ;
b) \({{1 - 3m} \over {2m}} + {{3m - 2} \over {2m - 1}} + {{3m - 2} \over {2m - 4{m^2}}}\) ;
c) \({k \over {{k^2} - 9}} + {1 \over {6k - 9 - {k^2}}} + {1 \over {{k^2} + 6k + 9}}\) .
Lời giải chi tiết
\(\eqalign{ & a)\,\,{4 \over {2 + y}} + {2 \over {y - 2}} + {{5y - 6} \over {4 - {y^2}}} \cr & \,\,\,\,\, = {4 \over {2 + y}} + {{ - 2} \over {2 - y}} + {{5y - 6} \over {\left( {2 - y} \right)\left( {2 + y} \right)}} \cr & \,\,\,\,\, = {{4\left( {2 - y} \right)} \over {\left( {2 + y} \right)\left( {2 - y} \right)}} + {{ - 2\left( {2 + y} \right)} \over {\left( {2 - y} \right)\left( {2 + y} \right)}} + {{5y - 6} \over {\left( {2 - y} \right)\left( {2 + y} \right)}} \cr & \,\,\,\,\, = {{8 - 4y - 4 - 2y + 5y - 6} \over {\left( {2 - y} \right)\left( {2 + y} \right)}} \cr & \,\,\,\,\, = {{ - y - 2} \over {\left( {2 - y} \right)\left( {2 + y} \right)}} = {{ - \left( {y + 2} \right)} \over {\left( {2 - y} \right)\left( {2 + y} \right)}} = {{ - 1} \over {2 - y}} \cr & b)\,\,{{1 - 3m} \over {2m}} + {{3m - 2} \over {2m - 1}} + {{3m - 2} \over {2m - 4{m^2}}} \cr & \,\,\,\,\, = {{1 - 3m} \over {2m}} + {{ - \left( {3m - 2} \right)} \over {1 - 2m}} + {{3m - 2} \over {2m\left( {1 - 2m} \right)}} \cr & \,\,\,\,\, = {{\left( {1 - 3m} \right)\left( {1 - 2m} \right)} \over {2m\left( {1 - 2m} \right)}} + {{ - \left( {3m - 2} \right).2m} \over {2m\left( {1 - 2m} \right)}} + {{3m - 2} \over {2m\left( {1 - 2m} \right)}} \cr & \,\,\,\,\, = {{1 - 2m - 3m + 6{m^2} - 6{m^2} + 4m + 3m - 2} \over {2m\left( {1 - 2m} \right)}} \cr & \,\,\,\,\, = {{2m - 1} \over {2m\left( {1 - 2m} \right)}} = {{ - \left( {1 - 2m} \right)} \over {2m\left( {1 - 2m} \right)}} = {{ - 1} \over {2m}} \cr & c)\,\,{k \over {{k^2} - 9}} + {1 \over {6k - 9 - {k^2}}} + {1 \over {{k^2} + 6k + 9}} \cr & \,\,\,\,\, = {k \over {\left( {k - 3} \right)\left( {k + 3} \right)}} + {{ - 1} \over {{{\left( {k - 3} \right)}^2}}} + {1 \over {{{\left( {k + 3} \right)}^2}}} \cr & \,\,\,\,\, = {{k\left( {k - 3} \right)\left( {k + 3} \right)} \over {{{\left( {k - 3} \right)}^2}{{\left( {k + 3} \right)}^2}}} + {{ - {{\left( {k + 3} \right)}^2}} \over {{{\left( {k - 3} \right)}^2}{{\left( {k + 3} \right)}^2}}} + {{{{\left( {k - 3} \right)}^2}} \over {{{\left( {k - 3} \right)}^2}{{\left( {k + 3} \right)}^2}}} \cr & \,\,\,\,\, = {{k\left( {{k^2} - 9} \right) - \left( {{k^2} + 6k + 9} \right) + \left( {{k^2} - 6k + 9} \right)} \over {{{\left( {k - 3} \right)}^2}{{\left( {k + 3} \right)}^2}}} \cr & \,\,\,\,\, = {{{k^3} - 9k - {k^2} - 6k - 9 + {k^2} - 6k + 9} \over {{{\left( {k - 3} \right)}^2}{{\left( {k + 3} \right)}^2}}} \cr & \,\,\,\,\, = {{{k^3} - 21k} \over {{{\left( {k - 3} \right)}^2}{{\left( {k + 3} \right)}^2}}} \cr} \)
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