Giải SBT Toán 8 KNTT Phép nhân và phép chia phân thức đại số có đáp án

\(\frac{{2{x^3}}}{{5{y^2}}}.\frac{{125{y^5}}}{{8x}} = \frac{{2{x^3}.125{y^5}}}{{5{y^2}.8x}}\) \( = \frac{{2{x^3}.5.25{y^5}}}{{5{y^2}.2.4x}} = \frac{{25{x^3}{y^5}}}{{4x{y^2}}} = \frac{{25{x^2}{y^3}}}{4}\).

\(\frac{{24{y^5}}}{{7{x^2}}}.\left( { - \frac{{21x}}{{12{y^3}}}} \right) = \frac{{24{y^5}.\left( { - 21x} \right)}}{{7{x^2}.12{y^3}}}\) \( = \frac{{12.2{y^5}.\left( { - 3} \right).7x}}{{7{x^2}.12{y^3}}} = \frac{{ - 6x{y^5}}}{{{x^2}{y^3}}} = \frac{{ - 6{y^2}}}{x}\).

\(\frac{{{x^2} - 6x + 9}}{{{x^2} - 3x + 9}}.\frac{{{x^3} + 27}}{{3x - 9}} = \frac{{{{\left( {x - 3} \right)}^2}}}{{{x^2} - 3x + 9}}.\frac{{\left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right)}}{{3\left( {x - 3} \right)}}\)

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

\(\frac{{2{x^2} - 20x + 50}}{{3x + 3}}.\frac{{{x^2} - 1}}{{4{{\left( {x - 5} \right)}^3}}}\)

\( = \frac{{2\left( {{x^2} - 10x + 25} \right)}}{{3\left( {x + 1} \right)}}.\frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{4{{\left( {x - 5} \right)}^3}}}\)

\( = \frac{{2{{\left( {x - 5} \right)}^2}}}{{3\left( {x + 1} \right)}}.\frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{4{{\left( {x - 5} \right)}^3}}}\)

\( = \frac{{2{{\left( {x - 5} \right)}^2}\left( {x - 1} \right)\left( {x + 1} \right)}}{{3\left( {x + 1} \right).4{{\left( {x - 5} \right)}^3}}} = \frac{{x - 1}}{{6\left( {x - 5} \right)}}\).

\(\frac{{{x^2} - {y^2}}}{{6{x^2}y}}:\frac{{x + y}}{{3xy}} = \frac{{\left( {x - y} \right)\left( {x + y} \right)}}{{6{x^2}y}}.\frac{{3xy}}{{x + y}}\)

\( = \frac{{\left( {x - y} \right)\left( {x + y} \right).3xy}}{{6{x^2}y.\left( {x + y} \right)}} = \frac{{x - y}}{{2x}}\).