Giải bài 6.22 trang 19 SGK Toán 8 tập 2 – Kết nối tri thức

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

\(\begin{array}{l}a)\frac{1}{x} + \frac{2}{{x + 1}} + \frac{3}{{x + 2}} – \frac{1}{x} – \frac{2}{{x – 1}} – \frac{3}{{x + 2}}\\ = \left( {\frac{1}{x} – \frac{1}{x}} \right) + \left( {\frac{2}{{x + 1}} – \frac{2}{{x – 1}}} \right) + \left( {\frac{3}{{x + 2}} – \frac{3}{{x + 2}}} \right)\\ = 0 + \frac{2}{{x + 1}} – \frac{2}{{x – 1}} + 0\\ = \frac{{2\left( {x – 1} \right) – 2\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x – 1} \right)}} = \frac{{2{\rm{x}} – 2 – 2{\rm{x}} – 2}}{{\left( {x + 1} \right)\left( {x – 1} \right)}} = \frac{{ – 4}}{{\left( {x + 1} \right)\left( {x – 1} \right)}}\end{array}\)

\(\begin{array}{l}b)\frac{{2{\rm{x}} – 1}}{x} + \frac{{1 – x}}{{2{\rm{x}} + 1}} + \frac{3}{{{x^2} – 9}} + \frac{{1 – 2{\rm{x}}}}{x} + \frac{{x – 1}}{{2{\rm{x}} + 1}} – \frac{3}{{x + 3}}\\ = \left( {\frac{{2{\rm{x}} – 1}}{x} + \frac{{1 – 2{\rm{x}}}}{x}} \right) + \left( {\frac{{1 – x}}{{2{\rm{x}} + 1}} + \frac{{x – 1}}{{2{\rm{x}} + 1}}} \right) + \left( {\frac{3}{{{x^2} – 9}} – \frac{3}{{x + 3}}} \right)\\ = 0 + 0 + \frac{3}{{\left( {x + 3} \right)\left( {x – 3} \right)}} – \frac{3}{{x + 3}}\\ = \frac{{3 – 3\left( {x – 3} \right)}}{{\left( {x + 3} \right)\left( {x – 3} \right)}} = \frac{{12 – 3{\rm{x}}}}{{\left( {x + 3} \right)\left( {x – 3} \right)}}\end{array}\)