Rút gọn:
1. \(A = {{a – b} \over {{a^2} + ab}}.{{{b^2} + ab} \over {{{\left( {a – b} \right)}^2}}}\)
2. \(B = \left( {{{{x^2} + {y^2}} \over {{x^2} – {y^2}}} – 1} \right).{{x – y} \over {2y}}.\)
3. \(C = {{{a^2} + 2ab + {b^2}} \over {{a^2} + ab + {b^2}}}.{{{a^3} – {b^3}} \over {7a + 7b}}\)
4. \(D = \left( {{{a + 1} \over {2a – 2}} – {1 \over {2{a^2} – 2}}} \right).{{2a + 2} \over {a + 2}}.\)
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1. \(A = {{a – b} \over {{a^2} + ab}}.{{{b^2} + ab} \over {{{\left( {a – b} \right)}^2}}} = {{\left( {a – b} \right)b\left( {a + b} \right)} \over {a\left( {a + b} \right){{\left( {a – b} \right)}^2}}} \)\(\;= {b \over {a\left( {a – b} \right)}}.\)
2. \(B = \left[ {{{{x^2} + {y^2} – \left( {{x^2} – {y^2}} \right)} \over {{x^2} – {y^2}}}} \right].{{x – y} \over {2y}} = {{2{y^2}\left( {x – y} \right)} \over {\left( {{x^2} – {y^2}} \right)2y}}\)\(\; = {y \over {x + y}}.\)
3. \(C = {{{{\left( {a + b} \right)}^2}\left( {a – b} \right)\left( {{a^2} + ab + {b^2}} \right)} \over {7\left( {{a^2} + ab + {b^2}} \right)\left( {a + b} \right)}} = {{{a^2} – {b^2}} \over 7}.\)
4.Ta có: \({{a + 1} \over {2a – 2}} – {1 \over {2{a^2} – 2}} = {{{{\left( {a + 1} \right)}^2} – 1} \over {2\left( {{a^2} – 1} \right)}} = {{{a^2} + 2a + 1 – 1} \over {2\left( {{a^2} – 1} \right)}} \)\(\;= {{a\left( {a + 2} \right)} \over {2\left( {{a^2} – 1} \right)}}\)
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Vậy \(D = {{a\left( {a + 2} \right)} \over {2\left( {{a^2} – 1} \right)}}.{{2\left( {a + 1} \right)} \over {a + 2}} = {a \over {a – 1}}.\)
