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Làm tính cộng các phân thức sau:
a)\({5 \over {2{x^2}y}} + {3 \over {5x{y^2}}} + {x \over {{y^3}}}\)
b)\({{x + 1} \over {2x + 6}} + {{2x + 3} \over {x\left( {x + 3} \right)}}\)
c)\({{3x + 5} \over {{x^2} – 5x}} + {{25 – x} \over {25 – 5x}}\)
d)\({x^2} + {{{x^4} + 1} \over {1 – {x^2}}} + 1)\)
e)\({{4{x^2} – 3x + 17} \over {{x^3} – 1}} + {{2x – 1} \over {{x^2} + x + 1}} + {6 \over {1 – x}}\)
Hướng dẫn làm bài:
a)\({5 \over {2{x^2}y}} + {3 \over {5x{y^2}}} + {x \over {{y^3}}} = {{5.5{y^2}} \over {2{x^2}y.5{y^2}}} + {{3.2xy} \over {5x{y^2}.2xy}} + {{x.10{x^2}} \over {{y^3}.10{x^2}}}\)
\( = {{25{y^2}} \over {10{x^2}{y^3}}} + {{6xy} \over {10{x^2}{y^3}}} + {{10{x^3}} \over {10{x^2}{y^3}}} = {{25{y^2} + 6xy + 10{x^3}} \over {10{x^2}{y^3}}}\)
b)\({{x + 1} \over {2x + 6}} + {{2x + 3} \over {x\left( {x + 3} \right)}} = {{x + 1} \over {2\left( {x + 3} \right)}} + {{2x + 3} \over {x\left( {x + 3} \right)}}\)
\( = {{x\left( {x + 1} \right)} \over {2x\left( {x + 3} \right)}} + {{2\left( {2x + 3} \right)} \over {2x\left( {x + 3} \right)}} = {{{x^2} + x + 4x + 6} \over {2x\left( {x + 3} \right)}}\)
\( = {{{x^2} + 5x + 6} \over {2x\left( {x + 3} \right)}} = {{{x^2} + 2x + 3x + 6} \over {2x\left( {x + 3} \right)}} = {{x\left( {x + 2} \right) + 3\left( {x + 2} \right)} \over {2x\left( {x + 3} \right)}}\)
c)\({{3x + 5} \over {{x^2} – 5x}} + {{25 – x} \over {25 – 5x}} = {{3x + 5} \over {{x^2} – 5x}} + {{x – 25} \over {5x – 25}}\)
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\( = {{3x + 5} \over {x\left( {x – 5} \right)}} + {{x – 25} \over {5\left( {x – 5} \right)}} = {{5\left( {3x + 5} \right)} \over {5x\left( {x – 5} \right)}} + {{x\left( {x – 25} \right)} \over {5x\left( {x – 5} \right)}}\)
\( = {{15x + 25 + {x^2} – 25x} \over {5x\left( {x – 5} \right)}} = {{{x^2} – 10x + 25} \over {5x\left( {x – 5} \right)}}\)
\( = {{{{\left( {x – 5} \right)}^2}} \over {5x\left( {x – 5} \right)}} = {{x – 5} \over {5x}}\)
d)\({x^2} + {{{x^4} + 1} \over {1 – {x^2}}} + 1 = 1 + {{\rm{x}}^2} + {{{x^4} + 1} \over {1 – {x^2}}}\)
\( = {{\left( {1 + {x^2}} \right)\left( {1 – {x^2}} \right)} \over {1 – {x^2}}} + {{{x^4} + 1} \over {1 – {x^2}}} = {{1 – {x^4} + {x^4} + 1} \over {1 – {x^2}}} = {2 \over {1 – {x^2}}}\)
e)\({{4{x^2} – 3x + 17} \over {{x^3} – 1}} + {{2x – 1} \over {{x^2} + x + 1}} + {6 \over {1 – x}}\)
\({{4{x^2} – 3x + 17} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} + {{2x – 1} \over {{x^2} + x + 1}} + {{ – 6} \over {x – 1}}\)
\( = {{4{x^2} – 3x + 17} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} + {{\left( {2x – 1} \right)\left( {x – 1} \right)} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} + {{ – 6\left( {{x^2} + x + 1} \right)} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\)
\( = {{4{x^2} – 3x + 17 + \left( {2x – 1} \right)\left( {x – 1} \right) – 6\left( {{x^2} + x + 1} \right)} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\)
\( = {{4{x^2} – 3x + 17 + 2{x^2} – 3x + 1 – 6{x^2} – 6x – 6} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\)
\( = {{ – 12x + 12} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} = {{ – 12\left( {x – 1} \right)} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} = {{ – 12} \over {{x^2} + x + 1}}\)