Thực hiện phép tính ( chú ý đến quy tắc đổi dấu)
a. \({{4\left( {x + 3} \right)} \over {3{x^2} – x}}:{{{x^2} + 3x} \over {1 – 3x}}\)
b. \({{4x + 6y} \over {x – 1}}:{{4{x^2} + 12xy + 9{y^2}} \over {1 – {x^3}}}\)
a. \({{4\left( {x + 3} \right)} \over {3{x^2} – x}}:{{{x^2} + 3x} \over {1 – 3x}}\)\( = {{4\left( {x + 3} \right)} \over {3{x^2} – x}}.{{1 – 3x} \over {{x^2} + 3x}} = {{4\left( {x + 3} \right)\left( {1 – 3x} \right)} \over {x\left( {3x – 1} \right).x\left( {x + 3} \right)}} = {{ – 4\left( {3x – 1} \right)} \over {{x^2}\left( {3x – 1} \right)}} = – {4 \over {{x^2}}}\)
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b. \({{4x + 6y} \over {x – 1}}:{{4{x^2} + 12xy + 9{y^2}} \over {1 – {x^3}}} = \)\({{4x + 6y} \over {x – 1}}.{{1 – {x^3}} \over {4{x^2} + 12xy + 9{y^2}}} = {{2\left( {2x + 3y} \right)\left( {1 – x} \right)\left( {1 + x + {x^2}} \right)} \over {\left( {x – 1} \right){{\left( {2x + 3y} \right)}^2}}}\)
\( = – {{2\left( {x – 1} \right)\left( {1 + x + {x^2}} \right)} \over {\left( {x – 1} \right)\left( {2x + 3y} \right)}} = – {{2\left( {1 + x + {x^2}} \right)} \over {2x + 3y}}\)