Giải bài 2 trang 42 SGK Toán 8 tập 1 - Cánh diều

Đề bài

Thực hiện phép tính:

\(a)\dfrac{{4x + 2}}{{4{{x  -  4}}}} + \dfrac{{3 - 6x}}{{6x - 6}}\)

\(b)\dfrac{y}{{2{x^2} - xy}} + \dfrac{{4x}}{{{y^2} - 2xy}}\)

\(c)\dfrac{x}{{x - y}} + \dfrac{y}{{x + y}} + \dfrac{{2{y^2}}}{{{x^2} - {y^2}}}\)

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

Lời giải chi tiết

\(\begin{array}{l}a)\dfrac{{4x + 2}}{{4{{x  -  4}}}} + \dfrac{{3 - 6x}}{{6x - 6}} \\ = \dfrac{{2\left( {2x + 1} \right)}}{{4\left( {x - 1} \right)}} + \dfrac{{3\left( {1 - 2x} \right)}}{{6\left( {x - 1} \right)}}\\ = \dfrac{{2x + 1}}{{2\left( {x - 1} \right)}} + \dfrac{{1 - 2x}}{{2\left( {x - 1} \right)}} \\ = \dfrac{{2x + 1 + 1 - 2x}}{{2\left( {x - 1} \right)}} \\ = \dfrac{2}{{2\left( {x - 1} \right)}} \\ = \dfrac{1}{{x - 1}}\end{array}\)

\(\begin{array}{l}b)\dfrac{y}{{2{x^2} - xy}} + \dfrac{{4x}}{{{y^2} - 2xy}} \\ = \dfrac{y}{{x\left( {2x - y} \right)}} + \dfrac{{4x}}{{y\left( {y - 2x} \right)}}\\ = \dfrac{y}{{x\left( {2x - y} \right)}} - \dfrac{{4x}}{{y\left( {2x - y} \right)}} \\ = \dfrac{{{y^2}}}{{xy\left( {2x - y} \right)}} - \dfrac{{4{x^2}}}{{xy\left( {2x - y} \right)}}\\ = \dfrac{{{y^2} - 4{x^2}}}{{xy\left( {2x - y} \right)}} \\ = \dfrac{{\left( {y - 2x} \right)\left( {y + 2x} \right)}}{{ - xy\left( {y - 2x} \right)}} \\ = \dfrac{{ - \left( {y + 2x} \right)}}{{xy}}\end{array}\)

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

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