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

Đề bài

Thực hiện phép tính:

\(a)\dfrac{{5{\rm{x}} - 4}}{9} + \dfrac{{4{\rm{x}} + 4}}{9}\)                                             b) \(\dfrac{{{x^2}y - 6}}{{2{{\rm{x}}^2}y}} + \dfrac{{6 - x{y^2}}}{{2{{\rm{x}}^2}y}}\)

c) \(\dfrac{{x + 1}}{{{x^2} - 5{\rm{x}}}} + \dfrac{{x - 16}}{{{x^2} - 5{\rm{x}}}} + \dfrac{{x + 2}}{{{x^2} - 5{\rm{x}}}}\)                           \(d)\dfrac{{7y}}{3} - \dfrac{{7y - 5}}{3}\)

e) \(\dfrac{{4{\rm{x}} - 1}}{{3{\rm{x}}{y^2}}} - \dfrac{{7{\rm{x}} - 1}}{{3{\rm{x}}{y^2}}}\)                                                 \(g)\dfrac{{3y - 2{\rm{x}}}}{{x - 2y}} - \dfrac{{x - y}}{{2y - x}}\)

Lời giải chi tiết

\(a)\dfrac{{5{\rm{x}} - 4}}{9} + \dfrac{{4{\rm{x}} + 4}}{9} = \dfrac{{5{\rm{x}} - 4 + 4{\rm{x}} + 4}}{9} = \dfrac{{9{\rm{x}}}}{9} = x\)

b) \(\dfrac{{{x^2}y - 6}}{{2{{\rm{x}}^2}y}} + \dfrac{{6 - x{y^2}}}{{2{{\rm{x}}^2}y}} = \dfrac{{{x^2}y - 6 + 6 - x{y^2}}}{{2{{\rm{x}}^2}y}} = \dfrac{{{x^2}y - x{y^2}}}{{2{{\rm{x}}^2}y}} = \dfrac{{xy\left( {x - y} \right)}}{{2{{\rm{x}}^2}y}} = \dfrac{{x - y}}{{2{\rm{x}}}}\)

c) \(\dfrac{{x + 1}}{{{x^2} - 5{\rm{x}}}} + \dfrac{{x - 16}}{{{x^2} - 5{\rm{x}}}} + \dfrac{{x + 2}}{{{x^2} - 5{\rm{x}}}} = \dfrac{{x + 1 + x - 16 + x + 2}}{{{x^2} - 5{\rm{x}}}} = \dfrac{{3{\rm{x}} - 15}}{{x\left( {x - 5} \right)}} = \dfrac{{3\left( {x - 5} \right)}}{{x\left( {x - 5} \right)}} = \dfrac{3}{x}\)

\(d)\dfrac{{7y}}{3} - \dfrac{{7y - 5}}{3} = \dfrac{{7y - 7y + 5}}{3} = \dfrac{5}{3}\)

e) \(\dfrac{{4{\rm{x}} - 1}}{{3{\rm{x}}{y^2}}} - \dfrac{{7{\rm{x}} - 1}}{{3{\rm{x}}{y^2}}} = \dfrac{{4{\rm{x}} - 1 - 7{\rm{x}} + 1}}{{3{\rm{x}}{y^2}}} = \dfrac{{3{\rm{x}}}}{{3{\rm{x}}{y^2}}} = \dfrac{1}{{{y^2}}}\)

\(g)\dfrac{{3y - 2{\rm{x}}}}{{x - 2y}} - \dfrac{{x - y}}{{2y - x}} = \dfrac{{3y - 2{\rm{x}}}}{{x - 2y}} + \left( { - \dfrac{{x - y}}{{2y - x}}} \right) = \dfrac{{3y - 2{\rm{x}}}}{{x - 2y}} + \dfrac{{x - y}}{{x - 2y}} = \dfrac{{3y - 2{\rm{x}} + x - y}}{{x - 2y}} = \dfrac{{2y - x}}{{ - \left( {2y - x} \right)}} =  - 1\)