Đề bài
Thực hiện phép nhân các phân thức:
a) \({{15{a^2}} \over {8a{b^3}c}}.{{4c} \over {5ab}}\) ;
b) \({{3(x + y)} \over {x - y}}.{{2x - 2y} \over {4x + 4y}}\) ;
c) \({{3{x^2}} \over {x - 5}}.{{{x^2} - 25} \over {9{x^3}}}\) .
Lời giải chi tiết
\(\eqalign{ & a)\,\,{{15{a^2}} \over {8a{b^3}c}}.{{4c} \over {5ab}} = {{15{a^2}.4c} \over {8a{b^3}c.5ab}} = {{60{a^2}c} \over {40{a^2}{b^4}c}} = {3 \over {2{b^4}}} \cr & b)\,\,{{3\left( {x + y} \right)} \over {x - y}}.{{2x - 2y} \over {4x + 4y}} = {{3\left( {x + y} \right)\left( {2x - 2y} \right)} \over {\left( {x - y} \right)\left( {4x + 4y} \right)}} = {{3\left( {x + y} \right)2\left( {x - y} \right)} \over {\left( {x - y} \right)4\left( {x + y} \right)}} = {3 \over 2} \cr & c)\,\,{{3{x^2}} \over {x - 5}}.{{{x^2} - 25} \over {9{x^3}}} = {{3{x^2}\left( {{x^2} - 25} \right)} \over {\left( {x - 5} \right)9{x^3}}} = {{3{x^2}\left( {x - 5} \right)\left( {x + 5} \right)} \over {\left( {x - 5} \right)3{x^2}.3x}} = {{x + 5} \over {3x}} \cr} \)
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