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Thực hiện các phép tính sau:
a) \(\left( {{{2x + 1} \over {2x – 1}} – {{2x – 1} \over {2x + 1}}} \right):{{4x} \over {10x – 5}}\)
b) \(\left( {{1 \over {{x^2} + x}} – {{2 – x} \over {x + 1}}} \right):\left( {{1 \over x} + x – 2} \right);\)
c) \({1 \over {x – 1}} – {{{x^3} – x} \over {{x^2} + 1}}.\left( {{1 \over {{x^2} – 2x + 1}} + {1 \over {1 – {x^2}}}} \right).\)
Hướng dẫn làm bài:
a) \(\left( {{{2x + 1} \over {2x – 1}} – {{2x – 1} \over {2x + 1}}} \right):{{4x} \over {10x – 5}} = {{{{\left( {2x + 1} \right)}^2} – {{\left( {2x – 1} \right)}^2}} \over {\left( {2x – 1} \right)\left( {2x + 1} \right)}}.{{10x + 5} \over {4x}}\)
=\({{4{x^2} + 4x + 1 – 4{x^2} + 4x – 1} \over {\left( {2x – 1} \right)\left( {2x + 1} \right)}}.{{5\left( {2x + 1} \right)} \over {4x}}\)
=\({{8x.5\left( {2x + 1} \right)} \over {\left( {2x – 1} \right)\left( {2x + 1} \right).4x}} = {{10} \over {2x – 1}}\)
b) \(\left( {{1 \over {{x^2} + x}} – {{2 – x} \over {x + 1}}} \right):\left( {{1 \over x} + x – 2} \right)\)
=\(\left( {{1 \over {x\left( {x + 1} \right)}} + {{x – 2} \over {x + 1}}} \right):{{1 + {x^2} – 2x} \over x}\)
=\({{1 + x\left( {x – 2} \right)} \over {x\left( {x + 1} \right)}}.{x \over {{x^2} – 2x + 1}}\)
=\({{\left( {{x^2} – 2x + 1} \right)x} \over {x\left( {x + 1} \right)\left( {{x^2} – 2x + 1} \right)}} = {1 \over {x + 1}}\)
c) \({1 \over {x – 1}} – {{{x^3} – x} \over {{x^2} + 1}}.\left( {{1 \over {{x^2} – 2x + 1}} + {1 \over {1 – {x^2}}}} \right)\)
=\({1 \over {x – 1}} – {{{x^3} – x} \over {{x^2} + 1}}.\left[ {{1 \over {{{\left( {x – 1} \right)}^2}}} – {1 \over {\left( {x – 1} \right)\left( {x + 1} \right)}}} \right]\)
=\({1 \over {x – 1}} – {{x\left( {{x^2} – 1} \right)} \over {{x^2} + 1}}.{{x + 1 – \left( {x – 1} \right)} \over {{{\left( {x – 1} \right)}^2}.\left( {x + 1} \right)}}\)
=\({1 \over {x – 1}} – {{x\left( {x – 1} \right)\left( {x + 1} \right)} \over {{x^2} + 1}}.{{x + 1 – x + 1} \over {{{\left( {x – 1} \right)}^2}\left( {x + 1} \right)}}\)
=\({1 \over {x – 1}} – {{x\left( {x – 1} \right)\left( {x + 1} \right).2} \over {\left( {{x^2} + 1} \right){{\left( {x – 1} \right)}^2}\left( {x + 1} \right)}} = {1 \over {x – 1}} – {{2x} \over {\left( {{x^2} + 1} \right)\left( {x – 1} \right)}}\)
=\({{{x^2} + 1 – 2x} \over {\left( {{x^2} + 1} \right)\left( {x – 1} \right)}} = {{{{\left( {x – 1} \right)}^2}} \over {\left( {{x^2} + 1} \right)\left( {x – 1} \right)}} = {{x – 1} \over {{x^2} + 1}}\)