Câu 36 trang 34 Sách bài tập (SBT) Toán 8 tập 1

Hãy làm các phép chia sau :

a. \({{7x + 2} \over {3x{y^3}}}:{{14x + 4} \over {{x^2}y}}\)

b. \({{8xy} \over {3x - 1}}:{{12x{y^3}} \over {5 - 15x}}\)

c. \({{27 - {x^3}} \over {5x + 5}}:{{2x - 6} \over {3x + 3}}\)

d. \(\left( {4{x^2} - 16} \right):{{3x + 6} \over {7x - 2}}\)

e. \({{3{x^3} + 3} \over {x - 1}}:\left( {{x^2} - x + 1} \right)\)

Giải:

a. \({{7x + 2} \over {3x{y^3}}}:{{14x + 4} \over {{x^2}y}}\)\( = {{7x + 2} \over {3x{y^3}}}.{{{x^2}y} \over {14x + 4}} = {{\left( {7x + 2} \right){x^2}y} \over {3x{y^3}.2\left( {7x + 2} \right)}} = {x \over {6{y^2}}}\)

b. \({{8xy} \over {3x - 1}}:{{12x{y^3}} \over {5 - 15x}}\)\( = {{8xy} \over {3x - 1}}.{{5 - 15x} \over {12x{y^3}}} = {{8xy\left( {5 - 15x} \right)} \over {\left( {3x - 1} \right).12x{y^3}}} = {{ - 10\left( {3x - 1} \right)} \over {3\left( {3x - 1} \right){y^2}}} = {{10} \over {3{y^2}}}\)

c. \({{27 - {x^3}} \over {5x + 5}}:{{2x - 6} \over {3x + 3}}\)\( = {{27 - {x^3}} \over {5x + 5}}:{{3x + 3} \over {2x - 6}} = {{\left( {{3^3} - {x^3}} \right).3\left( {x + 1} \right)} \over {5\left( {x + 1} \right).2\left( {x - 3} \right)}}\)

\( = {{ - 3\left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right)} \over {10\left( {x - 3} \right)}} =  - {{3\left( {{x^2} + 3x + 9} \right)} \over {10}}\)

d. \(\left( {4{x^2} - 16} \right):{{3x + 6} \over {7x - 2}}\)

\( = \left( {4{x^2} - 16} \right).{{7x - 2} \over {3x + 6}} = {{4\left( {x + 2} \right)\left( {x - 2} \right)\left( {7x - 2} \right)} \over {3\left( {x + 2} \right)}}\)

\( = {{4\left( {x - 2} \right)\left( {7x - 2} \right)} \over 3}\)

e. \({{3{x^3} + 3} \over {x - 1}}:\left( {{x^2} - x + 1} \right)\)\( = {{3{x^3} + 3} \over {x - 1}}.{1 \over {{x^2} - x + 1}} = {{3\left( {{x^3} + 1} \right)} \over {\left( {x - 1} \right)\left( {{x^2} - x + 1} \right)}} = {{3\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)} \over {\left( {x - 1} \right)\left( {{x^2} - x + 1} \right)}}\)

\( = {{3\left( {x + 1} \right)} \over {x - 1}}\)

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