Giải bài 3 trang 22 vở thực hành Toán 8 tập 2

Đề bài

Hãy thực hiện các phép tính đã chỉ ra.

a) \(\frac{{4{{\rm{x}}^2} – 1}}{{16{{\rm{x}}^2} – 1}}.\left( {\frac{1}{{2{\rm{x}} + 1}} + \frac{1}{{2{\rm{x}} – 1}} + \frac{1}{{1 – 4{{\rm{x}}^2}}}} \right)\);

b) \(\left( {\frac{{x + y}}{{xy}} – \frac{2}{x}} \right).\frac{{{x^3}{y^3}}}{{{x^3} – {y^3}}}\).

Lời giải chi tiết

a) Ta có: \(\frac{1}{{2{\rm{x}} + 1}} + \frac{1}{{2{\rm{x}} – 1}} = \frac{{2x – 1 + 2x + 1}}{{(2x + 1)(2x – 1)}} = \frac{{4x}}{{4{x^2} – 1}}\).

Do đó \(\frac{1}{{2{\rm{x}} + 1}} + \frac{1}{{2{\rm{x}} – 1}} + \frac{1}{{1 – 4{{\rm{x}}^2}}} = \frac{{4x}}{{4{x^2} – 1}} + \frac{{ – 1}}{{4{x^2} – 1}} = \frac{{4x – 1}}{{4{x^2} – 1}}\).

\(\begin{array}{l}\frac{{4{{\rm{x}}^2} – 1}}{{16{{\rm{x}}^2} – 1}}.\left( {\frac{1}{{2{\rm{x}} + 1}} + \frac{1}{{2{\rm{x}} – 1}} + \frac{1}{{1 – 4{{\rm{x}}^2}}}} \right) = \frac{{4{{\rm{x}}^2} – 1}}{{16{{\rm{x}}^2} – 1}}.\frac{{4{\rm{x}} – 1}}{{4{{\rm{x}}^2} – 1}}\\ = \frac{{\left( {4{{\rm{x}}^2} – 1} \right)\left( {4{\rm{x}} – 1} \right)}}{{\left( {4x – 1} \right)\left( {4x + 1} \right)\left( {4{{\rm{x}}^2} – 1} \right)}} = \frac{1}{{4x + 1}}.\end{array}\)

b) \(\left( {\frac{{x + y}}{{xy}} – \frac{2}{x}} \right).\frac{{{x^3}{y^3}}}{{{x^3} – {y^3}}} = \frac{{x – y}}{{xy}}.\frac{{{x^3}{y^3}}}{{\left( {x – y} \right)\left( {{x^2} + xy + {y^2}} \right)}} = \frac{{{x^2}{y^2}}}{{{x^2} + xy + {y^2}}}\).