Giải bài 1 trang 49 SGK Toán 8 tập 1 – Cánh diều

Đề bài

Thực hiện phép tính:

\(a)\dfrac{x}{{xy + {y^2}}} – \dfrac{y}{{{x^2} + xy}}\)                            

\(b)\dfrac{{{x^2} + 4}}{{{x^2} – 4}} – \dfrac{x}{{x + 2}} – \dfrac{x}{{2 – x}}\)

\(c)\dfrac{{{a^2} + ab}}{{b – a}}:\dfrac{{a + b}}{{2{{\rm{a}}^2} – 2{b^2}}}\)                          

\(d)\left( {\dfrac{{2{\rm{x}} + 1}}{{2{\rm{x}} – 1}} – \dfrac{{2{\rm{x}} – 1}}{{2{\rm{x}} + 1}}} \right):\dfrac{{4{\rm{x}}}}{{10{\rm{x}} – 5}}\)

Lời giải chi tiết

\(\begin{array}{l}a)\dfrac{x}{{xy + {y^2}}} – \dfrac{y}{{{x^2} + xy}}\\ = \dfrac{x}{{y\left( {x + y} \right)}} – \dfrac{y}{{x\left( {x + y} \right)}}\\ = \dfrac{{{x^2} – {y^2}}}{{xy\left( {x + y} \right)}} = \dfrac{{\left( {x – y} \right)\left( {x + y} \right)}}{{xy\left( {x + y} \right)}} = \dfrac{{x – y}}{{xy}}\end{array}\)

\(\begin{array}{l}b)\dfrac{{{x^2} + 4}}{{{x^2} – 4}} – \dfrac{x}{{x + 2}} – \dfrac{x}{{2 – x}}\\ = \dfrac{{{x^2} + 4}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} – \dfrac{x}{{x + 2}} + \dfrac{x}{{x – 2}}\\ = \dfrac{{{x^2} + 4 – x\left( {x – 2} \right) + x\left( {x + 2} \right)}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\ = \dfrac{{{x^2} + 4 – {x^2} + 2{\rm{x}} + {x^2} + 2{\rm{x}}}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = \dfrac{{{x^2} + 4{\rm{x}} + 4}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = \dfrac{{{{\left( {x + 2} \right)}^2}}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = \dfrac{{x + 2}}{{x – 2}}\end{array}\)

\(\begin{array}{l}c)\dfrac{{{a^2} + ab}}{{b – a}}:\dfrac{{a + b}}{{2{{\rm{a}}^2} – 2{b^2}}}\\ = \dfrac{{a\left( {a + b} \right)}}{{b – a}}.\dfrac{{2{{\rm{a}}^2} – 2{b^2}}}{{a + b}}\\ = \dfrac{{a\left( {a + b} \right).2.\left( {{a^2} – {b^2}} \right)}}{{ – \left( {a – b} \right).\left( {a + b} \right)}}\\ = \dfrac{{a\left( {a + b} \right).2.\left( {a – b} \right).\left( {a + b} \right)}}{{ – \left( {a – b} \right)\left( {a + b} \right)}} =  – 2{\rm{a}}\left( {a + b} \right)\end{array}\)

\(\begin{array}{l}d)\left( {\dfrac{{2{\rm{x}} + 1}}{{2{\rm{x}} – 1}} – \dfrac{{2{\rm{x}} – 1}}{{2{\rm{x}} + 1}}} \right):\dfrac{{4{\rm{x}}}}{{10{\rm{x}} – 5}}\\ = \dfrac{{{{\left( {2{\rm{x}} + 1} \right)}^2} – {{\left( {2{\rm{x}} – 1} \right)}^2}}}{{\left( {2{\rm{x}} + 1} \right)\left( {2{\rm{x}} – 1} \right)}}.\dfrac{{10x – 5}}{{4{\rm{x}}}}\\ = \dfrac{{\left( {2{\rm{x}} + 1 + 2{\rm{x}} – 1} \right)\left( {2{\rm{x}} + 1 – 2{\rm{x}} + 1} \right)}}{{\left( {2{\rm{x}} + 1} \right)\left( {2{\rm{x}} – 1} \right)}}.\dfrac{{5.\left( {2{\rm{x}} – 1} \right)}}{{4{\rm{x}}}}\\ = \dfrac{{4{\rm{x}}.2.5\left( {2{\rm{x}} – 1} \right)}}{{\left( {2{\rm{x}} + 1} \right)\left( {2{\rm{x}} – 1} \right).4{\rm{x}}}} = \dfrac{{10}}{{2{\rm{x}} + 1}}\end{array}\)