Giải Bài 2 trang 35 SGK Toán 8 tập 1 – Chân trời sáng tạo

Đề bài

Thực hiện các phép cộng, trừ phân thức sau:

a) \(\dfrac{1}{{2a}} + \dfrac{2}{{3b}}\)

b) \(\dfrac{{x - 1}}{{x + 1}} - \dfrac{{x + 1}}{{x - 1}}\)

c) \(\dfrac{{x + y}}{{xy}} - \dfrac{{y + z}}{{yz}}\)

d) \(\dfrac{2}{{x - 3}} - \dfrac{{12}}{{{x^2} - 9}}\)

e) \(\dfrac{1}{{x - 2}} + \dfrac{2}{{{x^2} - 4x + 4}}\)

Lời giải chi tiết

a) ĐKXĐ: \(a \ne 0;\;b \ne 0\)

\(\dfrac{1}{{2a}} + \dfrac{2}{{3b}}\) \( = \dfrac{{3b}}{{2a.3b}} + \dfrac{{2.2a}}{{3b.2a}} = \dfrac{{3b}}{{6ab}} + \dfrac{{4a}}{{6ab}} = \dfrac{{3b + 4a}}{{6ab}}\)

b) ĐKXĐ: \(x \ne  - 1;\;x \ne 1\)

\(\dfrac{{x - 1}}{{x + 1}} - \dfrac{{x + 1}}{{x - 1}}\) \( = \dfrac{{{{\left( {x - 1} \right)}^2}}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} - \dfrac{{{{\left( {x + 1} \right)}^2}}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} = \dfrac{{{x^2} - 2x + 1}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} - \dfrac{{{x^2} + 2x + 1}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} = \dfrac{{{x^2} - 2x + 1 - {x^2} - 2x - 1}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\)\( = \dfrac{{ - 4x}}{{{x^2} - 1}}\)

c) ĐKXĐ: \(x \ne 0;\;y \ne 0;\;z \ne 0\)

\(\dfrac{{x + y}}{{xy}} - \dfrac{{y + z}}{{yz}}\) \( = \dfrac{{\left( {x + y} \right).z}}{{xy.z}} - \dfrac{{\left( {y + z} \right).x}}{{yz.x}} = \dfrac{{xz + yz}}{{xyz}} - \dfrac{{xy + xz}}{{xyz}} = \dfrac{{xz + yz - xy - xz}}{{xyz}} = \dfrac{{yz - xy}}{{xyz}} = \dfrac{{y\left( {z - x} \right)}}{{xyz}} = \dfrac{{z - x}}{{xz}}\)

d) ĐKXĐ: \(x \ne  \pm 3\)

\(\dfrac{2}{{x - 3}} - \dfrac{{12}}{{{x^2} - 9}}\) \( = \dfrac{{2\left( {x + 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} - \dfrac{{12}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \dfrac{{2x + 6}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} - \dfrac{{12}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \dfrac{{2x - 6}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\)\( = \dfrac{{2\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\)\( = \dfrac{2}{{x + 3}}\)

e) ĐKXĐ: \(x \ne 2\)

\(\dfrac{1}{{x - 2}} + \dfrac{2}{{{x^2} - 4x + 4}}\) \( = \dfrac{{1.\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x - 2} \right)}} + \dfrac{2}{{{{\left( {x - 2} \right)}^2}}} = \dfrac{{x - 2 + 2}}{{{{\left( {x - 2} \right)}^2}}} = \dfrac{x}{{{{\left( {x - 2} \right)}^2}}}\)