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Làm tính nhân phân thức :
a. \({{3x – 2} \over {2xy}} – {{7x – 4} \over {2xy}}\)
b. \({{3x + 5} \over {4{x^3}y}} – {{5 – 15x} \over {4{x^3}y}}\)
c. \({{4x + 7} \over {2x + 2}} – {{3x + 6} \over {2x + 2}}\)
d. \({{9x + 5} \over {2\left( {x – 1} \right){{\left( {x + 3} \right)}^2}}} – {{5x – 7} \over {2\left( {x – 1} \right){{\left( {x + 3} \right)}^2}}}\)
e. \({{xy} \over {{x^2} – {y^2}}} – {{{x^2}} \over {{y^2} – {x^2}}}\)
f. \({{5x + {y^2}} \over {{x^2}y}} – {{5y – {x^2}} \over {x{y^2}}}\)
g. \({x \over {5x + 5}} – {x \over {10x – 10}}\)
h. \({{x + 9} \over {{x^2} – 9}} – {3 \over {{x^2} + 3x}}\)
a. \({{3x – 2} \over {2xy}} – {{7x – 4} \over {2xy}}\)\( = {{3x – 2} \over {2xy}} + {{4 – 7x} \over {2xy}} = {{3x – 2 + 4 – 7x} \over {2xy}} = {{2\left( {1 – 2x} \right)} \over {2xy}} = {{1 – 2x} \over {xy}}\)
b. \({{3x + 5} \over {4{x^3}y}} – {{5 – 15x} \over {4{x^3}y}}\)\( = {{3x + 5} \over {4{x^3}y}} + {{15x – 5} \over {4{x^3}y}} = {{3x + 5 + 15x – 5} \over {4{x^3}y}} = {{18x} \over {4{x^3}y}} = {9 \over {2{x^2}y}}\)
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c. \({{4x + 7} \over {2x + 2}} – {{3x + 6} \over {2x + 2}}\)\( = {{4x + 7} \over {2x + 2}} + {{ – \left( {3x + 6} \right)} \over {2x + 2}} = {{4x + 7 – 3x – 6} \over {2x + 2}} = {{x + 1} \over {2\left( {x + 1} \right)}} = {1 \over 2}\)
d. \({{9x + 5} \over {2\left( {x – 1} \right){{\left( {x + 3} \right)}^2}}} – {{5x – 7} \over {2\left( {x – 1} \right){{\left( {x + 3} \right)}^2}}}\)\( = {{9x + 5} \over {2\left( {x – 1} \right){{\left( {x + 3} \right)}^2}}} + {{7 – 5x} \over {2\left( {x – 1} \right){{\left( {x + 3} \right)}^2}}}\)
\( = {{9x + 5 + 7 – 5x} \over {2\left( {x – 1} \right){{\left( {x + 3} \right)}^2}}} = {{4\left( {x + 3} \right)} \over {2\left( {x – 1} \right){{\left( {x + 3} \right)}^2}}} = {2 \over {\left( {x – 1} \right)\left( {x + 3} \right)}}\)
e. \({{xy} \over {{x^2} – {y^2}}} – {{{x^2}} \over {{y^2} – {x^2}}}\)\( = {{xy} \over {{x^2} – {y^2}}} + {{{x^2}} \over {{x^2} – {y^2}}} = {{xy + {x^2}} \over {{x^2} – {y^2}}} = {{x\left( {x + y} \right)} \over {\left( {x + y} \right)\left( {x – y} \right)}} = {x \over {x – y}}\)
f. \({{5x + {y^2}} \over {{x^2}y}} – {{5y – {x^2}} \over {x{y^2}}}\)\( = {{5x + {y^2}} \over {{x^2}y}} + {{{x^2} – 5y} \over {x{y^2}}} = {{y\left( {5x + {y^2}} \right)} \over {{x^2}{y^2}}} + {{x\left( {{x^2} – 5y} \right)} \over {{x^2}{y^2}}}\)
\( = {{5xy + {y^3} + {x^3} – 5xy} \over {{x^2}{y^2}}} = {{{x^3} + {y^3}} \over {{x^2}{y^2}}}\)
g. \({x \over {5x + 5}} – {x \over {10x – 10}}\)\( = {x \over {5\left( {x + 1} \right)}} + {{ – x} \over {10\left( {x – 1} \right)}} = {{2x\left( {x – 1} \right)} \over {10\left( {x + 1} \right)\left( {x – 1} \right)}} + {{ – x\left( {x + 1} \right)} \over {10\left( {x + 1} \right)\left( {x – 1} \right)}}\)
\( = {{2{x^2} – 2x – {x^2} – x} \over {10\left( {x + 1} \right)\left( {x – 1} \right)}} = {{{x^2} – 3x} \over {10\left( {x + 1} \right)\left( {x – 1} \right)}}\)
h. \({{x + 9} \over {{x^2} – 9}} – {3 \over {{x^2} + 3x}}\)\( = {{x + 9} \over {\left( {x + 3} \right)\left( {x – 3} \right)}} + {{ – 3} \over {x\left( {x + 3} \right)}} = {{x\left( {x + 9} \right)} \over {x\left( {x + 3} \right)\left( {x – 3} \right)}} + {{ – 3\left( {x – 3} \right)} \over {x\left( {x + 3} \right)\left( {x – 3} \right)}}\)
\( = {{{x^2} + 9x – 3x + 9} \over {x\left( {x + 3} \right)\left( {x – 3} \right)}} = {{{x^2} + 6x + 9} \over {x\left( {x + 3} \right)\left( {x – 3} \right)}} = {{{{\left( {x + 3} \right)}^2}} \over {x\left( {x + 3} \right)\left( {x – 3} \right)}} = {{x + 3} \over {x\left( {x – 3} \right)}}\)