Đề bài

Thực hiện phép tính:

\(\begin{array}{l}a)\frac{{4{{{x}}^2} - 1}}{{16{{{x}}^2} - 1}}.\left( {\frac{1}{{2{{x}} + 1}} + \frac{1}{{2{{x}} - 1}} + \frac{1}{{1 - 4{{{x}}^2}}}} \right)\\b)\left( {\frac{{x + y}}{{xy}} - \frac{2}{x}} \right).\frac{{{x^3}{y^3}}}{{{x^3} - {y^3}}}\end{array}\)

Lời giải chi tiết

\(\begin{array}{l}a)\frac{{4{{{x}}^2} - 1}}{{16{{{x}}^2} - 1}}.\left( {\frac{1}{{2{{x}} + 1}} + \frac{1}{{2{{x}} - 1}} + \frac{1}{{1 - 4{{{x}}^2}}}} \right)\\ = \frac{{4{{{x}}^2} - 1}}{{16{{{x}}^2} - 1}}.\frac{{2{{x}} - 1 + 2{{x}} + 1 - 1}}{{\left( {2{{x}} - 1} \right)\left( {2{{x}} + 1} \right)}}\\ = \frac{{\left( {2{{x}} - 1} \right)\left( {2{{x}} + 1} \right)}}{{\left( {4{{x}} - 1} \right)\left( {4{{x + 1}}} \right)}}.\frac{{4{{x}} - 1}}{{\left( {2{{x}} - 1} \right)\left( {2{{x}} + 1} \right)}}\\ = \frac{1}{{4{{x}} + 1}}\\b)\left( {\frac{{x + y}}{{xy}} - \frac{2}{x}} \right).\frac{{{x^3}{y^3}}}{{{x^3} - {y^3}}}\\ = \frac{{x + y - 2y}}{{xy}}.\frac{{{x^3}{y^3}}}{{{x^3} - {y^3}}}\\ = \frac{{\left( {x - y} \right).{x^3}{y^3}}}{{xy\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)}} = \frac{{{x^2}{y^2}}}{{{x^2} + xy + y{}^2}}\end{array}\)