Bài 1. Thực hiện phép tính:
a) \(1:\left( {1 – {1 \over a}} \right)\)
b) \({{x – 2} \over {x – 5}}:\left( {{{{x^2} + 24} \over {{x^2} – 25}} – {4 \over {x – 5}}} \right).\)
Bài 2. Tìm P, biết: \({{a + 1} \over {{a^3} – 1}}.P = {{2a + 2} \over {{a^2} + a + 1}}.\)
Bài 3. Rút gọn: \(Q = \left( {{{{a^2} + {b^2}} \over a} + b} \right):\left[ {\left( {{1 \over {{a^2}}} + {1 \over {{b^2}}}} \right).{{{a^3} – {b^3}} \over {{a^2} + {b^2}}}} \right].\)
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Bài 1. a) \(1:\left( {1 – {1 \over a}} \right) = 1:{{a – 1} \over a} = 1.{a \over {a – 1}} = {a \over {a – 1}}.\)
b) \({{x – 2} \over {x – 5}}:\left( {{{{x^2} + 24} \over {{x^2} – 25}} – {4 \over {x – 5}}} \right) \)
\(= {{x – 2} \over {x – 5}}:{{{x^2} + 24 – 4x – 20} \over {{x^2} – 25}}\)
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\( = {{x – 2} \over {x – 5}}.{{{x^2} – 25} \over {{{\left( {x – 2} \right)}^2}}} = {{x + 5} \over {x – 2}}.\)
Bài 2. \(P = {{2a + 2} \over {{a^2} + a + 1}}:{{a + 1} \over {{a^3} – 1}} = {{2\left( {a + 1} \right)} \over {{a^2} + a + 1}}.{{{a^3} – 1} \over {a + 1}} \)\(\;= 2\left( {a – 1} \right).\)
Bài 3. \(Q = {{{a^2} + {b^2} + ab} \over a}:\left( {{{{a^2} + {b^2}} \over {{a^2}{b^2}}}.{{{a^3} – {b^3}} \over {{a^2} + {b^2}}}} \right) \)
\(\;\;\;\;= {{{a^2} + ab + {b^2}} \over a}:{{{a^3} – {b^3}} \over {{a^2}{b^2}}}\)
\( \;\;\;\;= {{{a^2} + ab + {b^2}} \over a}.{{{a^2}{b^2}} \over {{a^3} – {b^3}}} = {{a{b^2}} \over {a – b}}.\)