Câu 29 trang 32 Sách bài tập (SBT) Toán 8 tập 1

Làm tính nhân phân thức :

\(\eqalign{
& a.{{30{x^3}} \over {11{y^2}}}.{{121{y^5}} \over {25x}} \cr
& b.{{24{y^5}} \over {7{x^2}}}.\left( { - {{21x} \over {12{y^3}}}} \right) \cr
& c.\left( { - {{18{y^3}} \over {25{x^4}}}} \right).\left( { - {{15{x^2}} \over {9{y^3}}}} \right) \cr
& d.{{4x + 8} \over {{{\left( {x - 10} \right)}^3}}}.{{2x - 20} \over {{{\left( {x + 2} \right)}^2}}} \cr
& e.{{2{x^2} - 20x + 50} \over {3x + 3}}.{{{x^2} - 1} \over {4{{\left( {x - 5} \right)}^3}}} \cr} \)

Giải:

\(\eqalign{
& a)\;{{30{x^3}} \over {11{y^2}}}.{{121{y^5}} \over {25x}} = {{30{x^3}.121{y^5}} \over {11{y^2}.25x}} \cr
& = {{6{x^2}.11{y^3}} \over {1.5}} = {{66{x^2}{y^3}} \over 5} \cr} \)

\(\eqalign{
& b)\,\,{{24{y^5}} \over {7{x^2}}}.\left( { - {{21x} \over {12{y^3}}}} \right) = {{24{y^5}.\left( { - 21x} \right)} \over {7{x^2}.12{y^3}}} \cr
& = {{2{y^2}.\left( { - 3} \right)} \over x} = - {{6{y^2}} \over x} \cr} \)

\(\eqalign{
& c)\,\,\left( { - {{18{y^3}} \over {25{x^4}}}} \right).\left( { - {{15{x^2}} \over {9{y^3}}}} \right) \cr
& = {{\left( { - 18{y^3}} \right).\left( { - 15{x^2}} \right)} \over {25{x^4}.9{y^3}}} \cr
& = {{ - 2.\left( { - 3} \right)} \over {5{x^2}.1}} = {6 \over {5{x^2}}} \cr} \)

\(\eqalign{
& d)\,\,{{4x + 8} \over {{{\left( {x - 10} \right)}^3}}}.{{2x - 20} \over {{{\left( {x + 2} \right)}^2}}} \cr
& = {{4\left( {x + 2} \right).2\left( {x - 10} \right)} \over {{{\left( {x - 10} \right)}^3}{{\left( {x + 2} \right)}^2}}} \cr
& = {8 \over {{{\left( {x - 10} \right)}^2}\left( {x + 2} \right)}} \cr} \)

\(\eqalign{
& e)\,\,{{2{x^2} - 20x + 50} \over {3x + 3}}.{{{x^2} - 1} \over {4{{\left( {x - 5} \right)}^3}}} \cr
& = {{2\left( {{x^2} - 10x + 25} \right)\left( {x + 1} \right)\left( {x - 1} \right)} \over {3\left( {x + 1} \right).4{{\left( {x - 5} \right)}^3}}} \cr
& = {{{{\left( {x - 5} \right)}^2}\left( {x - 1} \right)} \over {6{{\left( {x - 5} \right)}^3}}} = {{x - 1} \over {6\left( {x - 5} \right)}} \cr} \)

Sachbaitap.com