Quy đồng mẫu thức các phân thức:
a) \( {{4b} \over {{b^2} – 2bc + {c^2}}};{{2a} \over {c – b}};{1 \over {4ac + 4ab}}\)
b) \( {x \over {{x^3} – 3{x^2} + 3x – 1}};{{3x} \over {{x^2} – 1}};{1 \over {{x^2} – 1}}\)
a) \( {b^2} – 2bc + {c^2} = {\left( {c – b} \right)^2}\)
\( 4ac + 4ab = 4a\left( {b + c} \right)\)
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\( MTC = 4a{\left( {c – b} \right)^2}\left( {b + c} \right)\)
Vậy: \( {{4b} \over {{b^2} – 2bc + {c^2}}} = {{16ab\left( {b + c} \right)} \over {4a{{\left( {c – b} \right)}^2}\left( {b + c} \right)}};\) \( {{2a} \over {c – b}} = {{8{a^2}\left( {{c^2} – {b^2}} \right)} \over {4a{{\left( {c – b} \right)}^2}\left( {b + c} \right)}};\)
\( {1 \over {4ac + 4ab}} = {{{{\left( {c – b} \right)}^2}} \over {4a{{\left( {c – b} \right)}^2}\left( {b + c} \right)}}\)
b) Ta có: \( {x^3} – 3{x^2} + 3x – 1 = {\left( {x – 1} \right)^3};\)
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\({x^2} – 1 = \left( {x – 1} \right)\left( {x + 1} \right)\)
\( {x^2} – x = x\left( {x – 1} \right)\)
\( MTC = x{\left( {x – 1} \right)^3}\left( {x + 1} \right)\)
Vậy: \( {x \over {{x^3} – 3{x^2} + 3x – 1}} = {{{x^2}\left( {x + 1} \right)} \over {x{{\left( {x – 1} \right)}^3}\left( {x + 1} \right)}};\)
\({{3x} \over {{x^2} – 1}} = {{3{x^2}{{\left( {x – 1} \right)}^2}} \over {x{{\left( {x – 1} \right)}^3}\left( {x + 1} \right)}};\)
\( {1 \over {{x^2} – x}} = {{{{\left( {x – 1} \right)}^2}\left( {x + 1} \right)} \over {x{{\left( {x – 1} \right)}^3}\left( {x + 1} \right)}}\)
