Câu 24 trang 30 Sách bài tập (SBT) Toán 8 tập 1

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. \({{x + 9} \over {{x^2} - 9}} - {3 \over {{x^2} + 3x}}\)

Giải:

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. \({{3x + 5} \over {4{x^3}y}} - {{5 - 15x} \over {4{x^3}y}}\)\( = {{3x + 5} \over {4{x^3}y}} + {{15x - 5} \over {4{x^3}y}} = {{3x + 5 + 15x - 5} \over {4{x^3}y}} = {{18x} \over {4{x^3}y}} = {9 \over {2{x^2}y}}\)

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. \({{x + 9} \over {{x^2} - 9}} - {3 \over {{x^2} + 3x}}\)\( = {{x + 9} \over {\left( {x + 3} \right)\left( {x - 3} \right)}} + {{ - 3} \over {x\left( {x + 3} \right)}} = {{x\left( {x + 9} \right)} \over {x\left( {x + 3} \right)\left( {x - 3} \right)}} + {{ - 3\left( {x - 3} \right)} \over {x\left( {x + 3} \right)\left( {x - 3} \right)}}\)

\( = {{{x^2} + 9x - 3x + 9} \over {x\left( {x + 3} \right)\left( {x - 3} \right)}} = {{{x^2} + 6x + 9} \over {x\left( {x + 3} \right)\left( {x - 3} \right)}} = {{{{\left( {x + 3} \right)}^2}} \over {x\left( {x + 3} \right)\left( {x - 3} \right)}} = {{x + 3} \over {x\left( {x - 3} \right)}}\)

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