Giải bài 3 trang 16 vở thực hành Toán 8 tập 2

Đề bài

Thực hiện các phép tính sau:

a) \(\frac{1}{x} + \frac{2}{{x + 1}} + \frac{3}{{x + 2}} - \frac{1}{x} - \frac{2}{{x + 1}} - \frac{3}{{x + 2}}\);

b) \(\frac{{2{\rm{x}} - 1}}{x} + \frac{{1 - x}}{{2{\rm{x}} + 1}} + \frac{3}{{{x^2} - 9}} + \frac{{1 - 2{\rm{x}}}}{x} + \frac{{x - 1}}{{2{\rm{x}} + 1}} - \frac{3}{{x + 3}}\).

Lời giải chi tiết

a)

\(\begin{array}{l}\frac{1}{x} + \frac{2}{{x + 1}} + \frac{3}{{x + 2}} - \frac{1}{x} - \frac{2}{{x - 1}} - \frac{3}{{x + 2}}\\ = \left( {\frac{1}{x} - \frac{1}{x}} \right) + \left( {\frac{2}{{x + 1}} - \frac{2}{{x - 1}}} \right) + \left( {\frac{3}{{x + 2}} - \frac{3}{{x + 2}}} \right)\\ = \frac{{2\left( {x - 1} \right) - 2\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} = \frac{{2{\rm{x}} - 2 - 2{\rm{x}} - 2}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} = \frac{{ - 4}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\end{array}\)

b)

\(\begin{array}{l}\frac{{2{\rm{x}} - 1}}{x} + \frac{{1 - x}}{{2{\rm{x}} + 1}} + \frac{3}{{{x^2} - 9}} + \frac{{1 - 2{\rm{x}}}}{x} + \frac{{x - 1}}{{2{\rm{x}} + 1}} - \frac{3}{{x + 3}}\\ = \left( {\frac{{2{\rm{x}} - 1}}{x} + \frac{{1 - 2{\rm{x}}}}{x}} \right) + \left( {\frac{{1 - x}}{{2{\rm{x}} + 1}} + \frac{{x - 1}}{{2{\rm{x}} + 1}}} \right) + \left( {\frac{3}{{{x^2} - 9}} - \frac{3}{{x + 3}}} \right)\\ = \frac{{3 - 3\left( {x - 3} \right)}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \frac{{3 - 3x + 9}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \frac{{12 - 3{\rm{x}}}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}\end{array}\)