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Thực hiện các phép tính sau :
a. \(\left( {{{5x + y} \over {{x^2} – 5xy}} + {{5x – y} \over {{x^2} + 5xy}}} \right).{{{x^2} – 25{y^2}} \over {{x^2} + {y^2}}}\)
b. \({{4xy} \over {{y^2} – {x^2}}}:\left( {{1 \over {{x^2} + 2xy + {y^2}}} – {1 \over {{x^2} – {y^2}}}} \right)\)
c. \(\left[ {{1 \over {{{\left( {2x – y} \right)}^2}}} + {2 \over {4{x^2} – {y^2}}} + {1 \over {{{\left( {2x + y} \right)}^2}}}} \right].{{4{x^2} + 4xy + {y^2}} \over {16x}}\)
d. \(\left( {{2 \over {x + 2}} – {4 \over {{x^2} + 4x + 4}}} \right):\left( {{2 \over {{x^2} – 4}} + {1 \over {2 – x}}} \right)\)
a. \(\left( {{{5x + y} \over {{x^2} – 5xy}} + {{5x – y} \over {{x^2} + 5xy}}} \right).{{{x^2} – 25{y^2}} \over {{x^2} + {y^2}}}\)
\(\eqalign{ & = \left[ {{{5x + y} \over {x\left( {x – 5y} \right)}} + {{5x – y} \over {x\left( {x + 5y} \right)}}} \right].{{{x^2} – 25{y^2}} \over {{x^2} + {y^2}}} \cr & = {{\left( {5x + y} \right)\left( {x + 5y} \right) + \left( {5x – y} \right)\left( {x – 5y} \right)} \over {x\left( {x – 5y} \right)\left( {x + 5y} \right)}}.{{\left( {x – 5y} \right)\left( {x + 5y} \right)} \over {{x^2} + {y^2}}} \cr & = {{5{x^2} + 25xy + xy + 5{y^2} + 5{x^2} – 25xy – xy + 5{y^2}} \over {x\left( {{x^2} + {y^2}} \right)}} \cr & = {{10{x^2} + 10{y^2}} \over {x\left( {{x^2} + {y^2}} \right)}} = {{10\left( {{x^2} + {y^2}} \right)} \over {x\left( {{x^2} + {y^2}} \right)}} = {{10} \over x} \cr} \)
b. \({{4xy} \over {{y^2} – {x^2}}}:\left( {{1 \over {{x^2} + 2xy + {y^2}}} – {1 \over {{x^2} – {y^2}}}} \right)\)
\(\eqalign{ & = {{4xy} \over {{y^2} – {x^2}}}:\left[ {{1 \over {{{\left( {x + y} \right)}^2}}} – {1 \over {\left( {x + y} \right)\left( {x – y} \right)}}} \right] \cr & = {{4xy} \over {{y^2} – {x^2}}}:{{x – y – \left( {x + y} \right)} \over {{{\left( {x + y} \right)}^2}\left( {x – y} \right)}} = {{4xy} \over {{y^2} – {x^2}}}:{{ – 2y} \over {{{\left( {x + y} \right)}^2}\left( {x – y} \right)}} = {{4xy} \over {{y^2} – {x^2}}}.{{{{\left( {x + y} \right)}^2}\left( {y – x} \right)} \over {2y}} \cr & = {{4xy{{\left( {x + y} \right)}^2}\left( {y – x} \right)} \over {\left( {y + x} \right)\left( {y – x} \right).2y}} = 2x\left( {x + y} \right) \cr} \)
c. \(\left[ {{1 \over {{{\left( {2x – y} \right)}^2}}} + {2 \over {4{x^2} – {y^2}}} + {1 \over {{{\left( {2x + y} \right)}^2}}}} \right].{{4{x^2} + 4xy + {y^2}} \over {16x}}\)
\(\eqalign{ & = \left[ {{1 \over {{{\left( {2x – y} \right)}^2}}} + {2 \over {\left( {2x + y} \right)\left( {2x – y} \right)}} + {1 \over {{{\left( {2x + y} \right)}^2}}}} \right].{{{{\left( {2x + y} \right)}^2}} \over {16x}} \cr & = {{{{\left( {2x + y} \right)}^2} + 2\left( {2x + y} \right)\left( {2x – y} \right) + {{\left( {2x – y} \right)}^2}} \over {{{\left( {2x + y} \right)}^2}.{{\left( {2x – y} \right)}^2}}}.{{{{\left( {2x + y} \right)}^2}} \over {16x}} \cr & = {{{{\left[ {\left( {2x + y} \right) + \left( {2x – y} \right)} \right]}^2}} \over {16x{{\left( {2x – y} \right)}^2}}} = {{{{\left( {4x} \right)}^2}} \over {16x{{\left( {2x – y} \right)}^2}}} = {{16{x^2}} \over {16x{{\left( {2x – y} \right)}^2}}} = {x \over {{{\left( {2x – y} \right)}^2}}} \cr} \)
d. \(\left( {{2 \over {x + 2}} – {4 \over {{x^2} + 4x + 4}}} \right):\left( {{2 \over {{x^2} – 4}} + {1 \over {2 – x}}} \right)\)
\(\eqalign{ & = \left[ {{2 \over {x + 2}} – {4 \over {{{\left( {x + 2} \right)}^2}}}} \right]:\left[ {{2 \over {\left( {x + 2} \right)\left( {x – 2} \right)}} – {1 \over {x – 2}}} \right] \cr & = {{2\left( {x + 2} \right) – 4} \over {{{\left( {x + 2} \right)}^2}}}:{{2 – \left( {x + 2} \right)} \over {\left( {x + 2} \right)\left( {x – 2} \right)}} = {{2x + 4 – 4} \over {{{\left( {x + 2} \right)}^2}}}:{{2 – x – 2} \over {\left( {x + 2} \right)\left( {x – 2} \right)}} \cr & = {{2x} \over {{{\left( {x + 2} \right)}^2}}}.{{\left( {x + 2} \right)\left( {x – 2} \right)} \over { – x}} = {{2\left( {x – 2} \right)} \over { – \left( {x + 2} \right)}} = {{2\left( {2 – x} \right)} \over {x + 2}} \cr} \)