Biến đổi các biểu thức sau thành phân thức:
a) \({{2x} \over {{1 \over x} + 1}}\) ; b) \({{1 + {2 \over {y – 1}}} \over {2y}}\)
c) \({{1 – {2 \over {x + 1}}} \over {1 + {{2 – {x^2}} \over {{x^2} – 1}}}}\) ; d) \({{{x \over 4} – 1 + {3 \over {4x}}} \over {{x \over 2} – {6 \over x} + {1 \over 2}}}\) .
\(\eqalign{ & a)\,\,{{2x} \over {{1 \over x} + 1}} = 2x:\left( {{1 \over x} + 1} \right) = 2x:{{1 + x} \over x} = 2x.{x \over {1 + x}} = {{2{x^2}} \over {1 + x}} \cr & b)\,\,{{1 + {2 \over {y – 1}}} \over {2y}} = \left( {1 + {2 \over {y – 1}}} \right):2y = {{y – 1 + 2} \over {y – 1}}.{1 \over {2y}} = {{y + 1} \over {y – 1}}.{1 \over {2y}} = {{y + 1} \over {2y\left( {y – 1} \right)}} \cr & c)\,{{1 – {2 \over {x + 1}}} \over {1 + {{2 – {x^2}} \over {{x^2} – 1}}}} = \left( {1 – {2 \over {x + 1}}} \right):\left( {1 + {{2 – {x^2}} \over {{x^2} – 1}}} \right) = {{x + 1 – 2} \over {x + 1}}:{{{x^2} – 1 + 2 – {x^2}} \over {{x^2} – 1}} \cr & = {{x – 1} \over {x + 1}}:{1 \over {{x^2} – 1}} = {{x – 1} \over {x + 1}}.{{{x^2} – 1} \over 1} = {{\left( {x – 1} \right)\left( {{x^2} – 1} \right)} \over {x + 1}} = {{\left( {x – 1} \right)\left( {x – 1} \right)\left( {x + 1} \right)} \over {x + 1}} = {\left( {x – 1} \right)^2} \cr & d)\,\,{{{x \over 4} – 1 + {3 \over {4x}}} \over {{x \over 2} – {6 \over x} + {1 \over 2}}} = \left( {{x \over 4} – 1 + {3 \over {4x}}} \right):\left( {{x \over 2} – {6 \over x} + {1 \over 2}} \right) = {{{x^2} – 4x + 3} \over {4x}}:{{{x^2} – 12 + x} \over {2x}} \cr & = {{{x^2} – 4x + 3} \over {4x}}.{{2x} \over {{x^2} – 12 + x}} = {{\left( {x – 1} \right)\left( {x – 3} \right).2x} \over {4x.\left( {x – 3} \right)\left( {x + 4} \right)}} = {{x – 1} \over {2\left( {x + 4} \right)}} \cr} \)