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Rút gọn biểu thức :
a) \({{3{a^2}b} \over {4c}}.{{{c^2}} \over {ab}}\) ;
b) \({9 \over {d – 3f}}:{6 \over {9f – 3d}}\) ;
c) \({{10k} \over {k – 5}}.{{3{k^2} – 15k} \over {5{k^2}}}\) ;
d) \({{4x – 1} \over {4y – 4x}}:{3 \over {9x – 9y}}\) ;
e) \({{mn – 4m + 2{n^2} – 8n} \over {m + 2n}}:{{4 – n} \over n}\) ;
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f) \({{7{{(p – q)}^2}} \over {pq}}.{{2{p^3}{q^2}} \over {p – {q^2}}}\)
\(\eqalign{ & a)\,\,{{3{a^2}b} \over {4c}}.{{{c^2}} \over {ab}} = {{3{a^2}b{c^2}} \over {4cab}} = {{3{a^2}b{c^2}} \over {4abc}} = {{3ac} \over 4} \cr & b)\,\,{9 \over {d – 3f}}:{6 \over {9f – 3d}} = {9 \over {d – 3f}}.{{9f – 3d} \over 6} \cr & \,\,\,\,\, = {9 \over {d – 3f}}.{{ – 3\left( {d – 3f} \right)} \over 6} = {{ – 9} \over 2} \cr & c)\,\,{{10k} \over {k – 5}}.{{3{k^2} – 15k} \over {5{k^2}}} = {{10k\left( {3{k^2} – 15k} \right)} \over {\left( {k – 5} \right)5{k^2}}} \cr & \,\,\,\,\, = {{10k.3k.\left( {k – 5} \right)} \over {\left( {k – 5} \right)5{k^2}}} = {{30{k^2}} \over {5{k^2}}} = 6 \cr & d)\,\,{{4x – 1} \over {4y – 4x}}:{3 \over {9x – 9y}} = {{ – \left( {4x – 1} \right)} \over {4x – 4y}}.{{9x – 9y} \over 3} \cr & \,\,\,\,\, = {{ – \left( {4x – 1} \right)9\left( {x – y} \right)} \over {4\left( {x – y} \right).3}} = {{ – \left( {4x – 1} \right).3} \over 4} = – {{3\left( {4x – 1} \right)} \over 4} \cr & e)\,\,{{mn – 4m + 2{n^2} – 8n} \over {m + 2n}}:{{4 – n} \over n} = {{mn – 4m + 2{n^2} – 8n} \over {m + 2n}}.{n \over {4 – n}} \cr & \,\,\,\,\,\, = {{m\left( {n – 4} \right) + 2n\left( {n – 4} \right)} \over {m + 2n}}.{n \over {4 – n}} \cr & \,\,\,\,\, = {{\left( {n – 4} \right)\left( {m + 2n} \right).n} \over {\left( {m + 2n} \right)\left( {4 – n} \right)}} = {{ – \left( {4 – n} \right)n} \over {4 – n}} = – n \cr & f)\,\,{{7{{\left( {p – q} \right)}^2}} \over {pq}}.{{2{p^3}{q^2}} \over {p – {q^2}}} = {{7{{\left( {p – q} \right)}^2}.2{p^3}{q^2}} \over {pq\left( {p – {q^2}} \right)}} = {{14{{\left( {p – q} \right)}^2}{p^2}q} \over {p – {q^2}}} \cr} \)