Câu 58: Thực hiện các phép tính :
a. \(\left( {{9 \over {{x^3} – 9x}} + {1 \over {x + 3}}} \right):\left( {{{x – 3} \over {{x^2} + 3x}} – {x \over {3x + 9}}} \right)\)
b. \(\left( {{2 \over {x – 2}} – {2 \over {x + 2}}} \right).{{{x^2} + 4x + 4} \over 8}\)
c. \(\left( {{{3x} \over {1 – 3x}} + {{2x} \over {3x + 1}}} \right):{{6{x^2} + 10x} \over {1 – 6x + 9{x^2}}}\)
d. \(\left( {{x \over {{x^2} – 25}} – {{x – 5} \over {{x^2} + 5x}}} \right):{{2x – 5} \over {{x^2} + 5x}} + {x \over {5 – x}}\)
e. \(\left( {{{{x^2} + xy} \over {{x^3} + {x^2}y + x{y^2} + {y^3}}} + {y \over {{x^2} + {y^2}}}} \right):\left( {{1 \over {x – y}} – {{2xy} \over {{x^3} – {x^2}y + x{y^2} – {y^3}}}} \right)\)
a. \(\left( {{9 \over {{x^3} – 9x}} + {1 \over {x + 3}}} \right):\left( {{{x – 3} \over {{x^2} + 3x}} – {x \over {3x + 9}}} \right)\)
\(\eqalign{ & = \left[ {{9 \over {x\left( {x + 3} \right)\left( {x – 3} \right)}} + {1 \over {x + 3}}} \right]:\left[ {{{x – 3} \over {x\left( {x + 3} \right)}} – {x \over {3\left( {x + 3} \right)}}} \right] \cr & = {{9 + x\left( {x – 3} \right)} \over {x\left( {x + 3} \right)\left( {x – 3} \right)}}:{{3\left( {x – 3} \right) – {x^2}} \over {3x\left( {x + 3} \right)}} = {{{x^2} – 3x + 9} \over {x\left( {x + 3} \right)\left( {x – 3} \right)}}.{{3x\left( {x + 3} \right)} \over {3x – 9 – {x^2}}} \cr & = {{3\left( {{x^2} – 3x + 9} \right)} \over {\left( {3 – x} \right)\left( {{x^2} – 3x + 9} \right)}} = {3 \over {3 – x}} \cr} \)
b. \(\left( {{2 \over {x – 2}} – {2 \over {x + 2}}} \right).{{{x^2} + 4x + 4} \over 8}\)\( = {{2\left( {x + 2} \right) – 2\left( {x – 2} \right)} \over {\left( {x – 2} \right)\left( {x + 2} \right)}}.{{{{\left( {x + 2} \right)}^2}} \over 8}\)
\( = {{2x + 4 – 2x + 4} \over {\left( {x – 2} \right)\left( {x + 2} \right)}}.{{{{\left( {x + 2} \right)}^2}} \over 8} = {8 \over {\left( {x – 2} \right)\left( {x + 2} \right)}}.{{{{\left( {x + 2} \right)}^2}} \over 8} = {{x + 2} \over {x – 2}}\)
c. \(\left( {{{3x} \over {1 – 3x}} + {{2x} \over {3x + 1}}} \right):{{6{x^2} + 10x} \over {1 – 6x + 9{x^2}}}\)\( = {{3x\left( {3x + 1} \right) + 2x\left( {1 – 3x} \right)} \over {\left( {1 – 3x} \right)\left( {1 + 3x} \right)}}:{{2x\left( {3x + 5} \right)} \over {{{\left( {1 – 3x} \right)}^2}}}\)
\(\eqalign{ & = {{9{x^2} + 3x + 2x – 6{x^2}} \over {\left( {1 – 3x} \right)\left( {1 + 3x} \right)}}.{{{{\left( {1 – 3x} \right)}^2}} \over {2x\left( {3x + 5} \right)}} = {{x\left( {3x + 5} \right)} \over {\left( {1 – 3x} \right)\left( {1 + 3x} \right)}}.{{{{\left( {1 – 3x} \right)}^2}} \over {2x\left( {3x + 5} \right)}} \cr & = {{1 – 3x} \over {2\left( {1 + 3x} \right)}} \cr} \)
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d. \(\left( {{x \over {{x^2} – 25}} – {{x – 5} \over {{x^2} + 5x}}} \right):{{2x – 5} \over {{x^2} + 5x}} + {x \over {5 – x}}\)
\(\eqalign{ & = \left[ {{x \over {\left( {x + 5} \right)\left( {x – 5} \right)}} – {{x – 5} \over {x\left( {x + 5} \right)}}} \right]:{{2x – 5} \over {x\left( {x + 5} \right)}} + {x \over {5 – x}} \cr & = {{{x^2} – {{\left( {x – 5} \right)}^2}} \over {x\left( {x + 5} \right)\left( {x – 5} \right)}}.{{x\left( {x + 5} \right)} \over {2x – 5}} + {x \over {5 – x}} \cr & = {{{x^2} – {x^2} + 10x – 25} \over {\left( {x – 5} \right)\left( {2x – 5} \right)}} + {x \over {5 – x}} = {{5\left( {2x – 5} \right)} \over {\left( {x – 5} \right)\left( {2x – 5} \right)}} – {x \over {x – 5}} \cr & = {5 \over {x – 5}} – {x \over {x – 5}} = {{5 – x} \over {x – 5}} = {{ – \left( {x – 5} \right)} \over {x – 5}} = – 1 \cr} \)
e. \(\left( {{{{x^2} + xy} \over {{x^3} + {x^2}y + x{y^2} + {y^3}}} + {y \over {{x^2} + {y^2}}}} \right):\left( {{1 \over {x – y}} – {{2xy} \over {{x^3} – {x^2}y + x{y^2} – {y^3}}}} \right)\)
\(\eqalign{ & = \left[ {{{{x^2} + xy} \over {\left( {{x^2} + {y^2}} \right)\left( {x + y} \right)}} + {y \over {{x^2} + {y^2}}}} \right]:\left[ {{1 \over {x – y}} – {{2xy} \over {\left( {{x^2} + {y^2}} \right)\left( {x – y} \right)}}} \right] \cr & = {{{x^2} + xy + y\left( {x + y} \right)} \over {\left( {{x^2} + {y^2}} \right)\left( {x + y} \right)}}:{{{x^2} + {y^2} – 2xy} \over {\left( {{x^2} + {y^2}} \right)\left( {x – y} \right)}} \cr & = {{{x^2} + xy + xy + {y^2}} \over {\left( {{x^2} + {y^2}} \right)\left( {x + y} \right)}}.{{\left( {{x^2} + {y^2}} \right)\left( {x – y} \right)} \over {{{\left( {x – y} \right)}^2}}} \cr & = {{{{\left( {x + y} \right)}^2}} \over {\left( {{x^2} + {y^2}} \right)\left( {x + y} \right)}}.{{\left( {{x^2} + {y^2}} \right)\left( {x – y} \right)} \over {{{\left( {x – y} \right)}^2}}} = {{x + y} \over {x – y}} \cr} \)
Câu 59: Chứng minh đẳng thức :
a. \(\left( {{{{x^2} – 2x} \over {2{x^2} + 8}} – {{2{x^2}} \over {8 – 4x + 2{x^2} – {x^3}}}} \right)\left( {1 – {1 \over x} – {2 \over {{x^2}}}} \right) = {{x + 1} \over {2x}}\)
b. \(\left[ {{2 \over {3x}} – {2 \over {x + 1}}.\left( {{{x + 1} \over {3x}} – x – 1} \right)} \right]:{{x – 1} \over x} = {{2x} \over {x – 1}}\)
c. \(\left[ {{2 \over {{{\left( {x + 1} \right)}^3}}}.\left( {{1 \over x} + 1} \right) + {1 \over {{x^2} + 2x + 1}}.\left( {{1 \over {{x^2}}} + 1} \right)} \right]:{{x – 1} \over {{x^3}}} = {x \over {x – 1}}\)
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a. Biến đổi vế trái :
\(\left( {{{{x^2} – 2x} \over {2{x^2} + 8}} – {{2{x^2}} \over {8 – 4x + 2{x^2} – {x^3}}}} \right)\left( {1 – {1 \over x} – {2 \over {{x^2}}}} \right)\)
\(\eqalign{ & = \left[ {{{{x^2} – 2x} \over {2\left( {{x^2} + 4} \right)}} – {{2{x^2}} \over {4\left( {2 – x} \right) + {x^2}\left( {2 – x} \right)}}} \right]{{{x^2} – x – 2} \over {{x^2}}} \cr & = \left[ {{{{x^2} – 2x} \over {2\left( {{x^2} + 4} \right)}} – {{2{x^2}} \over {\left( {2 – x} \right)\left( {4 + {x^2}} \right)}}} \right]{{{x^2} – x – 2} \over {{x^2}}} \cr & = {{\left( {{x^2} – 2x} \right)\left( {2 – x} \right) – 4{x^2}} \over {2\left( {2 – x} \right)\left( {{x^2} + 4} \right)}}.{{{x^2} – x – 2} \over {{x^2}}} \cr & = {{2{x^2} – {x^3} – 4x + 2{x^2} – 4{x^2}} \over {2\left( {2 – x} \right)\left( {{x^2} + 4} \right)}}.{{{x^2} – 2x + x – 2} \over {{x^2}}} \cr & = {{ – x\left( {{x^2} + 4} \right)} \over {2\left( {2 – x} \right)\left( {{x^2} + 4} \right)}}.{{x\left( {x – 2} \right) + \left( {x – 2} \right)} \over {{x^2}}} \cr & = {{x\left( {{x^2} + 4} \right)} \over {2\left( {x – 2} \right)\left( {{x^2} + 4} \right)}}.{{\left( {x – 2} \right)\left( {x + 1} \right)} \over {{x^2}}} = {{x + 1} \over {2x}} \cr} \)
Vế trái bằng vế phải, vậy đẳng thức được chứng minh.
b. Biến đổi vế trái:
\(\eqalign{ & \left[ {{2 \over {3x}} – {2 \over {x + 1}}.\left( {{{x + 1} \over {3x}} – x – 1} \right)} \right]:{{x – 1} \over x} \cr & = \left[ {{2 \over {3x}} – {2 \over {x + 1}}.{{x + 1 – 3x\left( {x + 1} \right)} \over {3x}}} \right].{x \over {x – 1}} \cr & = \left[ {{2 \over {3x}} – {2 \over {x + 1}}.{{\left( {x + 1} \right)\left( {1 – 3x} \right)} \over {3x}}} \right].{x \over {x – 1}} \cr & = \left[ {{2 \over {3x}} – {{2\left( {1 – 3x} \right)} \over {3x}}} \right].{x \over {x – 1}} = {{2 – 2 + 6x} \over {3x}}.{x \over {x – 1}} = 2.{x \over {x – 1}} = {{2x} \over {x – 1}} \cr} \)
Vế trái bằng vế phải, vậy đẳng thức được chứng minh.
c. Biến đổi vế trái :
\(\eqalign{ & \left[ {{2 \over {{{\left( {x + 1} \right)}^3}}}.\left( {{1 \over x} + 1} \right) + {1 \over {{x^2} + 2x + 1}}.\left( {{1 \over {{x^2}}} + 1} \right)} \right]:{{x – 1} \over {{x^3}}} \cr & = \left[ {{2 \over {{{\left( {x + 1} \right)}^3}}}.{{x + 1} \over x} + {1 \over {{{\left( {x + 1} \right)}^2}}}.{{{x^2} + 1} \over {{x^2}}}} \right].{{{x^3}} \over {x – 1}} \cr & = \left[ {{2 \over {x{{\left( {x + 1} \right)}^2}}} + {{{x^2} + 1} \over {{x^2}{{\left( {x + 1} \right)}^2}}}} \right].{{{x^3}} \over {x – 1}} = {{2x + {x^2} + 1} \over {{x^2}{{\left( {x + 1} \right)}^2}}}.{{{x^3}} \over {x – 1}} \cr & = {{{{\left( {x + 1} \right)}^2}} \over {{x^2}{{\left( {x + 1} \right)}^2}}}.{{{x^3}} \over {x – 1}} = {x \over {x – 1}} \cr} \)
Vế trái bằng vế phải, vậy đẳng thức được chứng minh.
Câu 60: Biến đổi các biểu thức hữu tỉ thành phân thức :
a. \({{{x \over {x – 1}} – {{x + 1} \over x}} \over {{x \over {x + 1}} – {{x – 1} \over x}}}\)
b. \({{{5 \over 4} – {5 \over {x + 1}}} \over {{{9 – {x^2}} \over {{x^2} + 2x + 1}}}}\)
a. \({{{x \over {x – 1}} – {{x + 1} \over x}} \over {{x \over {x + 1}} – {{x – 1} \over x}}}\)\( = \left( {{x \over {x – 1}} – {{x + 1} \over x}} \right):\left( {{x \over {x + 1}} – {{x – 1} \over x}} \right)\)
\( = {{{x^2} – \left( {x + 1} \right)\left( {x – 1} \right)} \over {x\left( {x – 1} \right)}}:{{{x^2} – \left( {x – 1} \right)\left( {x + 1} \right)} \over {x\left( {x + 1} \right)}} = {1 \over {x\left( {x – 1} \right)}}.{{x\left( {x + 1} \right)} \over 1} = {{x + 1} \over {x – 1}}\)
b. \({{{5 \over 4} – {5 \over {x + 1}}} \over {{{9 – {x^2}} \over {{x^2} + 2x + 1}}}}\)\( = \left( {{5 \over 4} – {5 \over {x + 1}}} \right):\left( {{{9 – {x^2}} \over {{x^2} + 2x + 1}}} \right) = {{5\left( {x + 1} \right) – 20} \over {4\left( {x + 1} \right)}}:{{\left( {3 + x} \right)\left( {3 – x} \right)} \over {{{\left( {x + 1} \right)}^2}}}\)
\( = {{5\left( {x – 3} \right)} \over {4\left( {x + 1} \right)}}.{{{{\left( {x + 1} \right)}^2}} \over {\left( {3 + x} \right)\left( {3 – x} \right)}} = {{ – 5\left( {3 – x} \right)\left( {x + 1} \right)} \over {4\left( {3 + x} \right)\left( {3 – x} \right)}} = {{ – 5\left( {x + 1} \right)} \over {4\left( {3 + x} \right)}}\)
