Nội dung bài giảng
Biến đổi các biểu thức sau thành phân thức
a. \({1 \over 2} + {x \over {1 - {x \over {x + 2}}}}\)
b. \({{x - {1 \over {{x^2}}}} \over {x + {1 \over x} + {1 \over {{x^2}}}}}\)
c. \({{1 - {{2y} \over x} + {{{y^2}} \over {{x^2}}}} \over {{1 \over x} - {1 \over y}}}\)
d. \({{{x \over 4} - 1 + {3 \over {4x}}} \over {{x \over 2} - {6 \over x} + {1 \over 2}}}\)
Giải:
a. \({1 \over 2} + {x \over {1 - {x \over {x + 2}}}}\)\( = {1 \over 2} + {x \over {{{x + 2 - x} \over {x + 2}}}} = {1 \over 2} + {x \over {{2 \over {x + 2}}}}\)
b. \({{x - {1 \over {{x^2}}}} \over {x + {1 \over x} + {1 \over {{x^2}}}}}\) \( = \left( {x - {1 \over {{x^2}}}} \right):\left( {1 + {1 \over x} + {1 \over {{x^2}}}} \right) = {{{x^3} - 1} \over {{x^2}}}:{{{x^2} + x + 1} \over {{x^2}}}\)
\( = {{{x^3} - 1} \over {{x^2}}}.{{{x^2}} \over {{x^2} + x + 1}} = {{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right){x^2}} \over {{x^2}\left( {{x^2} + x + 1} \right)}} = x - 1\)
c. \({{1 - {{2y} \over x} + {{{y^2}} \over {{x^2}}}} \over {{1 \over x} - {1 \over y}}}\)\( = \left( {1 - {{2y} \over x} + {{{y^2}} \over {{x^2}}}} \right):\left( {{1 \over x} - {1 \over y}} \right) = {{{x^2} - 2xy + {y^2}} \over {{x^2}}}:{{y - x} \over {xy}}\)
\( = {{{x^2} - 2xy + {y^2}} \over {{x^2}}}.{{xy} \over {y - x}} = {{{{\left( {y - x} \right)}^2}.xy} \over {{x^2}\left( {y - x} \right)}} = {{y\left( {y - x} \right)} \over x}\)
d. \({{{x \over 4} - 1 + {3 \over {4x}}} \over {{x \over 2} - {6 \over x} + {1 \over 2}}}\)\( = \left( {{x \over 4} - 1 + {3 \over {4x}}} \right):\left( {{x \over 2} - {6 \over x} + {1 \over 2}} \right) = {{{x^2} - 4x + 3} \over {4x}}:{{{x^2} - 12x + x} \over {2x}}\)
\(\eqalign{ & = {{{x^2} - 4x + 3} \over {4x}}.{{2x} \over {{x^2} - 12 + x}} = {{{x^2} - x - 3x + 3} \over {4x}}.{{2x} \over {{x^2} - 3x + 4x - 12}} \cr & = {{\left( {x - 1} \right)\left( {x - 3} \right)} \over {4x}}.{{2x} \over {\left( {x - 3} \right)\left( {x + 4} \right)}} = {{\left( {x - 1} \right)\left( {x - 3} \right).2x} \over {4x\left( {x - 3} \right)\left( {x + 4} \right)}} = {{x - 1} \over {2\left( {x + 4} \right)}} \cr} \)