Giải bài 3 trang 42 SGK Toán 8 tập 1 - Cánh diều

Đề bài

Thực hiện phép tính:

\(a)\dfrac{1}{{x - 2}} - \dfrac{1}{{x + 1}}\)                                 b) \(\dfrac{{12}}{{{x^2} - 9}} - \dfrac{2}{{x - 3}}\)

c) \(\dfrac{1}{{xy - {x^2}}} - \dfrac{1}{{{y^2} - xy}}\)                       d) \(\dfrac{{2{\rm{x}}}}{{{x^2} - 1}} - \dfrac{3}{{2 + 2{\rm{x}}}} + \dfrac{1}{{2 - 2{\rm{x}}}}\)

Lời giải chi tiết

\(\begin{array}{l}a)\dfrac{1}{{x - 2}} - \dfrac{1}{{x + 1}}\\ = \dfrac{{x + 1}}{{\left( {x - 2} \right)\left( {x + 1} \right)}} - \dfrac{{x - 2}}{{\left( {x - 2} \right)\left( {x + 1} \right)}} = \dfrac{{x + 1 - x + 2}}{{\left( {x - 2} \right)\left( {x + 1} \right)}} = \dfrac{3}{{\left( {x - 2} \right)\left( {x + 1} \right)}}\end{array}\)

b) \(\begin{array}{l}\dfrac{{12}}{{{x^2} - 9}} - \dfrac{2}{{x - 3}}\\ = \dfrac{{12}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} - \dfrac{2}{{x - 3}}\\ = \dfrac{{12}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} - \dfrac{{2\left( {x + 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \dfrac{{12 - 2{\rm{x}} - 6}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \dfrac{{6 - 2{\rm{x}}}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \dfrac{{ - 2\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \dfrac{{ - 2}}{{x + 3}}\end{array}\)

c) \(\begin{array}{l}\dfrac{1}{{xy - {x^2}}} - \dfrac{1}{{{y^2} - xy}}\\ = \dfrac{1}{{x\left( {y - x} \right)}} - \dfrac{1}{{y\left( {y - x} \right)}}\\ = \dfrac{y}{{xy\left( {y - x} \right)}} - \dfrac{x}{{xy\left( {y - x} \right)}}\\ = \dfrac{{y - x}}{{xy\left( {y - x} \right)}} = \dfrac{1}{{xy}}\end{array}\)

d) \(\begin{array}{l}\dfrac{{2{\rm{x}}}}{{{x^2} - 1}} - \dfrac{3}{{2 + 2{\rm{x}}}} + \dfrac{1}{{2 - 2{\rm{x}}}}\\ = \dfrac{{2{\rm{x}}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} - \dfrac{3}{{2\left( {x + 1} \right)}} - \dfrac{1}{{2{\rm{x}} - 2}}\\ = \dfrac{{2{\rm{x}}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} - \dfrac{{3\left( {x - 1} \right)}}{{2\left( {x + 1} \right)\left( {x - 1} \right)}} - \dfrac{1}{{2\left( {x - 1} \right)}}\\ = \dfrac{{{\rm{4x}}}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}} - \dfrac{{3\left( {x - 1} \right)}}{{2\left( {x + 1} \right)\left( {x - 1} \right)}} - \dfrac{{1\left( {x + 1} \right)}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\ = \dfrac{{{\rm{4x}} - 3{\rm{x}} + 3 - x - 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{2}{{2\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\end{array}\)