Giải SBT Toán 8 KNTT Phép cộng và phép trừ phân thức đại số có đáp án

\(\frac{{{x^2} - 2}}{{x{{\left( {x - 1} \right)}^2}}} + \frac{{2 - x}}{{x{{\left( {x - 1} \right)}^2}}}\)

\( = \frac{{{x^2} - 2 + 2 - x}}{{x{{\left( {x - 1} \right)}^2}}}\)\( = \frac{{{x^2} - x}}{{x{{\left( {x - 1} \right)}^2}}}\)

\( = \frac{{x\left( {x - 1} \right)}}{{x{{\left( {x - 1} \right)}^2}}} = \frac{1}{{x - 1}}\).

\(\frac{{1 - 2x}}{{6{x^3}y}} + \frac{{3 + 2x}}{{6{x^3}y}} + \frac{{2x - 4}}{{6{x^3}y}}\)

=\(\frac{{1 - 2x + 3 + 2x + 2x - 4}}{{6{x^3}y}}\)

=\(\frac{{2x}}{{6{x^3}y}} = \frac{1}{{3{x^2}y}}\).

\(\frac{{2{x^2} - 1}}{{{x^2} - 3x}} - \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{{x^2} - 3x}}\)

= \(\frac{{2{x^2} - 1}}{{{x^2} - 3x}} - \frac{{{x^2} - 1}}{{{x^2} - 3x}}\)

= \(\frac{{2{x^2} - 1 - \left( {{x^2} - 1} \right)}}{{{x^2} - 3x}}\)

= \(\frac{{2{x^2} - 1 - {x^2} + 1}}{{{x^2} - 3x}}\)

= \(\frac{{{x^2}}}{{{x^2} - 3x}}\)

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

\(\frac{1}{{2x - 3}} - \frac{{13}}{{\left( {2x - 3} \right)\left( {4x + 7} \right)}}\)

= \(\frac{{4x + 7}}{{\left( {2x - 3} \right)\left( {4x + 7} \right)}} - \frac{{13}}{{\left( {2x - 3} \right)\left( {4x + 7} \right)}}\) (Mẫu thức chung là: (2x – 3)(4x + 7))

= \(\frac{{4x + 7 - 13}}{{\left( {2x - 3} \right)\left( {4x + 7} \right)}}\)

= \(\frac{{4x - 6}}{{\left( {2x - 3} \right)\left( {4x + 7} \right)}}\)

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

= \(\frac{2}{{4x + 7}}\).

\(\frac{{5x + {y^2}}}{{{x^2}y}} - \frac{{5y - {x^2}}}{{x{y^2}}}\)

\( = \frac{{y\left( {5x + {y^2}} \right)}}{{{x^2}{y^2}}} - \frac{{x\left( {5y - {x^2}} \right)}}{{{x^2}{y^2}}}\) (Mẫu thức chung là: x²y²)

\( = \frac{{5xy + {y^3}}}{{{x^2}{y^2}}} - \frac{{5xy - {x^3}}}{{{x^2}{y^2}}}\)

\( = \frac{{5xy + {y^3} - 5xy + {x^3}}}{{{x^2}{y^2}}} = \frac{{{x^3} + {y^3}}}{{{x^2}{y^2}}}\).

\(\frac{y}{{2{x^2} - xy}} + \frac{{4x}}{{{y^2} - 2xy}}\)=\(\frac{y}{{x\left( {2x - y} \right)}} + \frac{{4x}}{{y\left( {y - 2x} \right)}}\)

=\(\frac{{ - {y^2}}}{{xy\left( {y - 2x} \right)}} + \frac{{4{x^2}}}{{xy\left( {y - 2x} \right)}}\)           (Mẫu thức chung là: xy(y – 2x))

= \(\frac{{ - {y^2} + 4{x^2}}}{{xy\left( {y - 2x} \right)}}\)= \(\frac{{{{\left( {2x} \right)}^2} - {y^2}}}{{xy\left( {y - 2x} \right)}}\) = \(\frac{{\left( {2x - y} \right)\left( {2x + y} \right)}}{{xy\left( {y - 2x} \right)}}\)

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

\(\frac{5}{{6{x^2}y}} + \frac{7}{{12x{y^2}}} + \frac{{11}}{{18xy}}\)

= \(\frac{{30y}}{{36{x^2}{y^2}}} + \frac{{21x}}{{36{x^2}{y^2}}} + \frac{{22xy}}{{36{x^2}{y^2}}}\)       (Mẫu thức chung là: 36x²y²)

= \(\frac{{30y + 21x + 22xy}}{{36{x^2}{y^2}}}\).

\(\frac{{{x^3} + 2x}}{{{x^3} + 1}} + \frac{{2x}}{{{x^2} - x + 1}} + \frac{1}{{x + 1}}\)

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

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

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

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

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

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

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