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