Câu 29: Làm tính nhân phân thức :
a. \({{30{x^3}} \over {11{y^2}}}.{{121{y^5}} \over {25x}}\)
b. \({{24{y^5}} \over {7{x^2}}}.\left( { – {{21x} \over {12{y^3}}}} \right)\)
c. \(\left( { – {{18{y^3}} \over {25{x^4}}}} \right).\left( { – {{15{x^2}} \over {9{y^3}}}} \right)\)
d. \({{4x + 8} \over {{{\left( {x – 10} \right)}^3}}}.{{2x – 20} \over {{{\left( {x + 2} \right)}^2}}}\)
e. \({{2{x^2} – 20x + 50} \over {3x + 3}}.{{{x^2} – 1} \over {4{{\left( {x – 5} \right)}^3}}}\)
a. \({{30{x^3}} \over {11{y^2}}}.{{121{y^5}} \over {25x}}\)\( = {{30{x^3}.121{y^5}} \over {11{y^2}.25x}} = {{6{x^2}.11{y^3}} \over {1.5}} = {{66{x^2}{y^3}} \over 5}\)
b. \({{24{y^5}} \over {7{x^2}}}.\left( { – {{21x} \over {12{y^3}}}} \right)\) \( = {{24{y^5}.\left( { – 21x} \right)} \over {7{x^2}.12{y^3}}} = {{2{y^2}.\left( { – 3} \right)} \over x} = – {{6{y^2}} \over x}\)
c. \(\left( { – {{18{y^3}} \over {25{x^4}}}} \right).\left( { – {{15{x^2}} \over {9{y^3}}}} \right)\) \( = {{\left( { – 18{y^3}} \right).\left( { – 15{x^2}} \right)} \over {25{x^4}.9{y^3}}} = {{ – 2.\left( { – 3} \right)} \over {5{x^2}.1}} = {6 \over {5{x^2}}}\)
d. \({{4x + 8} \over {{{\left( {x – 10} \right)}^3}}}.{{2x – 20} \over {{{\left( {x + 2} \right)}^2}}}\)\( = {{4\left( {x + 2} \right).2\left( {x – 10} \right)} \over {{{\left( {x – 10} \right)}^3}{{\left( {x + 2} \right)}^2}}} = {8 \over {{{\left( {x – 10} \right)}^2}\left( {x + 2} \right)}}\)
e. \({{2{x^2} – 20x + 50} \over {3x + 3}}.{{{x^2} – 1} \over {4{{\left( {x – 5} \right)}^3}}}\)\( = {{2\left( {{x^2} – 10x + 25} \right)\left( {x + 1} \right)\left( {x – 1} \right)} \over {3\left( {x + 1} \right).4{{\left( {x – 5} \right)}^3}}}\)
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\( = {{{{\left( {x – 5} \right)}^2}\left( {x – 1} \right)} \over {6{{\left( {x – 5} \right)}^3}}} = {{x – 1} \over {6\left( {x – 5} \right)}}\)
Câu 30: Rút gọn biểu thức (chú ý dùng quy tắc đổi dấu để thấy nhân tử chung) :
a. \({{x + 3} \over {{x^2} – 4}}.{{8 – 12x + 6{x^2} – {x^3}} \over {9x + 27}}\)
b. \({{6x – 3} \over {5{x^2} + x}}.{{25{x^2} + 10x + 1} \over {1 – 8{x^3}}}\)
c. \({{3{x^2} – x} \over {{x^2} – 1}}.{{1 – {x^4}} \over {{{\left( {1 – 3x} \right)}^3}}}\)
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a. \({{x + 3} \over {{x^2} – 4}}.{{8 – 12x + 6{x^2} – {x^3}} \over {9x + 27}}\)\({{\left( {x + 3} \right)\left( {8 – 12x + 6{x^2} – {x^3}} \right)} \over {\left( {x + 2} \right)\left( {x – 2} \right).9\left( {x + 3} \right)}}\)
\( = {{{2^3} – {{3.2}^2}.x + 3.2{x^2} – {x^3}} \over {9\left( {x + 2} \right)\left( {x – 2} \right)}} = {{{{\left( {2 – x} \right)}^3}} \over { – 9\left( {x + 2} \right)\left( {2 – x} \right)}} = – {{{{\left( {2 – x} \right)}^2}} \over {9\left( {x + 2} \right)}}\)
b. \({{6x – 3} \over {5{x^2} + x}}.{{25{x^2} + 10x + 1} \over {1 – 8{x^3}}}\)\( = {{3\left( {2x – 1} \right){{\left( {5x + 1} \right)}^2}} \over {x\left( {5x + 1} \right)\left[ {1 – {{\left( {2x} \right)}^2}} \right]}} = {{3\left( {2x – 1} \right)\left( {5x + 1} \right)} \over {x\left( {1 – 2x} \right)\left( {1 + 2x + 4{x^2}} \right)}}\)
\( = – {{3\left( {2x – 1} \right)\left( {5x + 1} \right)} \over {x\left( {2x – 1} \right)\left( {1 + 2x + 4{x^2}} \right)}} = – {{3\left( {5x + 1} \right)} \over {x\left( {1 + 2x + 4{x^2}} \right)}}\)
c. \({{3{x^2} – x} \over {{x^2} – 1}}.{{1 – {x^4}} \over {{{\left( {1 – 3x} \right)}^3}}}\)\( = {{x\left( {3x – 1} \right)\left( {1 – {x^4}} \right)} \over {\left( {{x^2} – 1} \right){{\left( {1 – 3x} \right)}^3}}} = {{x\left( {3x – 1} \right)\left( {{x^2} – 1} \right)\left( {{x^2} + 1} \right)} \over {\left( {{x^2} – 1} \right){{\left( {3x – 1} \right)}^3}}}\)
\( = {{x\left( {{x^2} + 1} \right)} \over {{{\left( {3x – 1} \right)}^2}}}\)
Câu 31: Phân tích các tử thức và các mẫu thức (nếu cần thì dùng phương pháp thêm và bớt cùng một số hạng hoặc tách một số hạng thành hai số hạng) rồi rút gọn biểu thức :
a. \({{x – 2} \over {x + 1}}.{{{x^2} – 2x – 3} \over {{x^2} – 5x + 6}}\)
b. \({{x + 1} \over {{x^2} – 2x – 8}}.{{4 – x} \over {{x^2} + x}}\)
c. \({{x + 2} \over {4x + 24}}.{{{x^2} – 36} \over {{x^2} + x – 2}}\)
a. \({{x – 2} \over {x + 1}}.{{{x^2} – 2x – 3} \over {{x^2} – 5x + 6}}\)\( = {{\left( {x – 2} \right)\left( {{x^2} – 2x – 3} \right)} \over {\left( {x + 1} \right)\left( {{x^2} – 5x + 6} \right)}} = {{\left( {x – 2} \right)\left( {{x^2} – 3x + x – 3} \right)} \over {\left( {x + 1} \right)\left( {{x^2} – 2x – 3x + 6} \right)}}\)
\( = {{\left( {x – 2} \right)\left[ {x\left( {x – 3} \right) + \left( {x – 3} \right)} \right]} \over {\left( {x + 1} \right)\left[ {x\left( {x – 2} \right) – 3\left( {x – 2} \right)} \right]}} = {{\left( {x – 2} \right)\left( {x – 3} \right)\left( {x + 1} \right)} \over {\left( {x + 1} \right)\left( {x – 2} \right)\left( {x – 3} \right)}} = 1\)
b. \({{x + 1} \over {{x^2} – 2x – 8}}.{{4 – x} \over {{x^2} + x}}\)\( = {{\left( {x + 1} \right)\left( {4 – x} \right)} \over {\left( {{x^2} – 2x – 8} \right)x\left( {x + 1} \right)}} = {{4 – x} \over {\left( {{x^2} – 4x + 2x – 8} \right)x}}\)
\( = {{4 – x} \over {\left[ {x\left( {x – 4} \right) + 2\left( {x – 4} \right)} \right]x}} = {{4 – x} \over {x\left( {x – 4} \right)\left( {x + 2} \right)}} = – {{x – 4} \over {x\left( {x – 4} \right)\left( {x + 2} \right)}} = – {1 \over {x\left( {x + 2} \right)}}\)
c. \({{x + 2} \over {4x + 24}}.{{{x^2} – 36} \over {{x^2} + x – 2}}\)\({{\left( {x + 2} \right)\left( {x + 6} \right)\left( {x – 6} \right)} \over {4\left( {x + 6} \right)\left( {{x^2} + x – 2} \right)}} = {{\left( {x + 2} \right)\left( {x – 6} \right)} \over {4\left( {{x^2} + 2x – x – 2} \right)}}\)
\( = {{\left( {x + 2} \right)\left( {x – 6} \right)} \over {4\left[ {x\left( {x + 2} \right) – \left( {x – 2} \right)} \right]}} = {{\left( {x + 2} \right)\left( {x – 6} \right)} \over {4\left( {x + 2} \right)\left( {x – 1} \right)}} = {{x – 6} \over {4\left( {x – 1} \right)}}\)
