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Thực hiện các phép chia phân thức:
a) \({{a – 2b} \over {16}}:{{2a – 4b} \over {12}}\) ;
b) \({{3(c + d)} \over {c – d}}:{{12c + 12d} \over {3c – 3d}}\) ;
c) \({{a – 2b} \over 8}:{{4a – 8b} \over {12}}\) ;
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d) \({{8{c^3}} \over {6(c + d)}}:{{2{c^2}} \over {3c + 3d}}\) .
\(\eqalign{ & a)\,\,{{a – 2b} \over {16}}:{{2a – 4b} \over {12}} = {{a – 2b} \over {16}}.{{12} \over {2a – 4b}} = {{\left( {a – 2b} \right).12} \over {16\left( {2a – 4b} \right)}} \cr & = {{\left( {a – 2b} \right).12} \over {32\left( {a – 2b} \right)}} = {3 \over 8} \cr & b)\,\,{{3\left( {c + d} \right)} \over {c – d}}:{{12c + 12d} \over {3c – 3d}} = {{3\left( {c + d} \right)} \over {c – d}}.{{3c – 3d} \over {12c + 12d}} \cr & = {{3\left( {c + d} \right)\left( {3c – 3d} \right)} \over {\left( {c – d} \right)\left( {12c + 12d} \right)}} = {{3\left( {c + d} \right)3\left( {c – d} \right)} \over {\left( {c – d} \right)12\left( {c + d} \right)}} = {3 \over 4} \cr & c)\,\,{{a – 2b} \over 8}:{{4a – 8b} \over {12}} = {{a – 2b} \over 8}.{{12} \over {4a – 8b}} \cr & = {{\left( {a – 2b} \right).12} \over {8\left( {4a – 8b} \right)}} = {{\left( {a – 2b} \right).12} \over {32\left( {a – 2b} \right)}} = {3 \over 8} \cr & d)\,\,{{8{c^3}} \over {6\left( {c + d} \right)}}:{{2{c^2}} \over {3c + 3d}} = {{8{c^3}} \over {6\left( {c + d} \right)}}.{{3c + 3d} \over {2{c^2}}} \cr & = {{8{c^3}.\left( {3c + 3d} \right)} \over {6\left( {c + d} \right).2{c^2}}} = {{8{c^3}.3\left( {c + d} \right)} \over {6\left( {c + d} \right).2{c^2}}} = {3 \over 4} \cr} \)