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