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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. \({{3{x^2} – x} \over {{x^2} – 1}}.{{1 – {x^4}} \over {{{\left( {1 – 3x} \right)}^3}}}\)
a. \({{x + 3} \over {{x^2} – 4}}.{{8 – 12x + 6{x^2} – {x^3}} \over {9x + 27}}\)\({{\left( {x + 3} \right)\left( {8 – 12x + 6{x^2} – {x^3}} \right)} \over {\left( {x + 2} \right)\left( {x – 2} \right).9\left( {x + 3} \right)}}\)
\( = {{{2^3} – {{3.2}^2}.x + 3.2{x^2} – {x^3}} \over {9\left( {x + 2} \right)\left( {x – 2} \right)}} = {{{{\left( {2 – x} \right)}^3}} \over { – 9\left( {x + 2} \right)\left( {2 – x} \right)}} = – {{{{\left( {2 – x} \right)}^2}} \over {9\left( {x + 2} \right)}}\)
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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. \({{3{x^2} – x} \over {{x^2} – 1}}.{{1 – {x^4}} \over {{{\left( {1 – 3x} \right)}^3}}}\)\( = {{x\left( {3x – 1} \right)\left( {1 – {x^4}} \right)} \over {\left( {{x^2} – 1} \right){{\left( {1 – 3x} \right)}^3}}} = {{x\left( {3x – 1} \right)\left( {{x^2} – 1} \right)\left( {{x^2} + 1} \right)} \over {\left( {{x^2} – 1} \right){{\left( {3x – 1} \right)}^3}}}\)
\( = {{x\left( {{x^2} + 1} \right)} \over {{{\left( {3x – 1} \right)}^2}}}\)