Rút gọn phân thức
Bài 1. \({{{a^2}} \over {a – b}} + {{{b^2}} \over {b – a}}\)
Bài 2. \(4 + {{3a} \over {5 – 2b}} + {{5\left( {a – 10} \right)} \over {2b – 5}}\)
Bài 3. \({{3{x^2} – x + 3} \over {{x^3} – 1}} + {{1 – x} \over {{x^2} + x + 1}} + {2 \over {1 – x}}\)
Bài 1. \({{{a^2}} \over {a – b}} + {{{b^2}} \over {b – a}} = {{{a^2}} \over {a – b}} + {{ – {b^2}} \over {a – b}} = {{{a^2} – {b^2}} \over {a – b}}\)\(\; = a + b\)
Bài 2.
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\(4 + {{3a} \over {5 – 2b}} + {{5\left( {a – 10} \right)} \over {2b – 5}} \)
\(= 4 + {{3a} \over {5 – 2b}} + {{ – 5\left( {a – 10} \right)} \over {5 – 2b}}\)
\( = {{4\left( {5 – 2b} \right) + 3a – 5\left( {a – 10} \right)} \over {5 – 2b}} \)
\(= {{20 – 8b + 3a – 5a + 50} \over {5 – 2b}}\)
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\( = {{70 – 8b – 2a} \over {5 – 2b}}\)
Bài 3 \(MTC = {x^3} – 1 = \left( {{x^2} + x + 1} \right)\left( {x – 1} \right)\)
\({{3{x^2} – x + 3} \over {{x^3} – 1}} + {{1 – x} \over {{x^2} + x + 1}} + {2 \over {1 – x}} \)
\(= {{3{x^2} – x + 3} \over {{x^3} – 1}} + {{1 – x} \over {{x^2} + x + 1}} + {{ – 2} \over {x – 1}}\)
\( = {{3{x^2} – x + 3 + \left( {1 – x} \right)\left( {x – 1} \right) – 2\left( {{x^2} + x + 1} \right)} \over {{x^3} – 1}}\)
\( = {{3{x^2} – x + 3 – {x^2} + 2x – 1 – 2{x^2} – 2x – 1} \over {{x^3} – 1}} = {{ – x + 1} \over {{x^3} – 1}}\)
\( = {{ – \left( {x – 1} \right)} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} = {{ – 1} \over {{x^2} + x + 1}}\)
