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Rút gọn các phân thức sau:
a. \({{14x{y^5}\left( {2x – 3y} \right)} \over {21{x^2}y{{\left( {2x – 3y} \right)}^2}}}\)
b. \({{8xy{{\left( {3x – 1} \right)}^3}} \over {12{x^3}\left( {1 – 3x} \right)}}\)
c. \({{20{x^2} – 45} \over {{{\left( {2x + 3} \right)}^2}}}\)
d.\({{5{x^2} – 10xy} \over {2{{\left( {2y – x} \right)}^3}}}\)
e. \({{80{x^3} – 125x} \over {3\left( {x – 3} \right) – \left( {x – 3} \right)\left( {8 – 4x} \right)}}\)
f. \({{9 – {{\left( {x + 5} \right)}^2}} \over {{x^2} + 4x + 4}}\)
g. \({{32x – 8{x^2} + 2{x^3}} \over {{x^3} + 64}}\)
h. \({{5{x^3} + 5x} \over {{x^4} – 1}}\)
i. \({{{x^2} + 5x + 6} \over {{x^2} + 4x + 4}}\)
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a. \({{14x{y^5}\left( {2x – 3y} \right)} \over {21{x^2}y{{\left( {2x – 3y} \right)}^2}}}\) \(= {{2{y^4}} \over {3x\left( {2x – 3y} \right)}}\)
b. \({{8xy{{\left( {3x – 1} \right)}^3}} \over {12{x^3}\left( {1 – 3x} \right)}}\) \( = {{ – 8xy{{\left( {3x – 1} \right)}^3}} \over {12{x^2}\left( {3x – 1} \right)}} = {{ – 2y{{\left( {3x – 1} \right)}^2}} \over {3x}}\)
c. \({{20{x^2} – 45} \over {{{\left( {2x + 3} \right)}^2}}}\) \( = {{5\left( {4{x^2} – 9} \right)} \over {{{\left( {2x + 3} \right)}^2}}} = {{5\left( {2x + 3} \right)\left( {2x – 3} \right)} \over {{{\left( {2x + 3} \right)}^2}}} = {{5\left( {2x – 3} \right)} \over {2x + 3}}\)
d. \({{5{x^2} – 10xy} \over {2{{\left( {2y – x} \right)}^3}}}\) \( = {{ – 5x\left( {2y – x} \right)} \over {2{{\left( {2y – x} \right)}^3}}} = {{ – 5x} \over {2{{\left( {2y – x} \right)}^2}}}\)
e. \({{80{x^3} – 125x} \over {3\left( {x – 3} \right) – \left( {x – 3} \right)\left( {8 – 4x} \right)}}\) \( = {{5x\left( {16{x^2} – 25} \right)} \over {\left( {x – 3} \right)\left( {3 – 8 + 4x} \right)}} = {{5x\left( {16{x^2} – 25} \right)} \over {\left( {x – 3} \right)\left( {4x – 5} \right)}} = {{5x\left( {4x + 5} \right)} \over {x – 3}}\)
f. \({{9 – {{\left( {x + 5} \right)}^2}} \over {{x^2} + 4x + 4}}\) \( = {{\left( {3 + x + 5} \right)\left( {3 – x – 5} \right)} \over {{{\left( {x + 2} \right)}^2}}} = {{ – \left( {8 + x} \right)\left( {x + 2} \right)} \over {{{\left( {x + 2} \right)}^2}}} = {{ – \left( {8 + x} \right)} \over {x + 2}}\)
g. \({{32x – 8{x^2} + 2{x^3}} \over {{x^3} + 64}}\) \( = {{2x\left( {16 – 4x + {x^2}} \right)} \over {\left( {x + 4} \right)\left( {{x^2} – 4x + 16} \right)}} = {{2x} \over {x + 4}}\)
h. \({{5{x^3} + 5x} \over {{x^4} – 1}}\)\( = {{5x\left( {{x^2} + 1} \right)} \over {\left( {{x^2} – 1} \right)\left( {{x^2} + 1} \right)}} = {{5x} \over {{x^2} – 1}}\)
i. \({{{x^2} + 5x + 6} \over {{x^2} + 4x + 4}}\) \( = {{{x^2} + 2x + 3x + 6} \over {{{\left( {x + 2} \right)}^2}}} = {{x\left( {x + 2} \right) + 3\left( {x + 2} \right)} \over {{{\left( {x + 2} \right)}^2}}} = {{\left( {x + 2} \right)\left( {x + 3} \right)} \over {{{\left( {x + 2} \right)}^2}}} = {{x + 3} \over {x + 2}}\)