Câu hỏi 5 - Mục Bài tập trang 81

1. Nội dung câu hỏi

Thực hiện các phép tính sau:

a) \(\frac{{2x + 4}}{{x + 3}} + \frac{3}{x} - \frac{6}{{{x^2} + 3x}}\);

b) \(\frac{{{x^2} - 3x - 4}}{{{x^2} + x}}:\frac{{{x^2} - x - 12}}{{{x^2} - x}}\).

3. Lời giải chi tiết

a) \(\frac{{2x + 4}}{{x + 3}} + \frac{3}{x} - \frac{6}{{{x^2} + 3x}} = \frac{{x\left( {2x + 4} \right)}}{{x\left( {x + 3} \right)}} + \frac{{3\left( {x + 3} \right)}}{{x\left( {x + 3} \right)}} - \frac{6}{{x\left( {x + 3} \right)}}\)\( = \frac{{2{x^2} + 4x + 3x + 9 - 6}}{{x\left( {x + 3} \right)}}\)

\( = \frac{{2{x^2} + 7x + 3}}{{x\left( {x + 3} \right)}} = \frac{{2{x^2} + 6x + x + 3}}{{x\left( {x + 3} \right)}} = \frac{{2x\left( {x + 3} \right) + \left( {x + 3} \right)}}{{x\left( {x + 3} \right)}} = \frac{{\left( {x + 3} \right)\left( {2x + 1} \right)}}{{x\left( {x + 3} \right)}} = \frac{{2x + 1}}{x}\)

b) \(\frac{{{x^2} - 3x - 4}}{{{x^2} + x}}:\frac{{{x^2} - x - 12}}{{{x^2} - x}} = \frac{{{x^2} + x - 4x - 4}}{{x\left( {x + 1} \right)}}:\frac{{{x^2} - 4x + 3x - 12}}{{x\left( {x - 1} \right)}}\)

\( = \frac{{x\left( {x + 1} \right) - 4\left( {x + 1} \right)}}{{x\left( {x + 1} \right)}}:\frac{{x\left( {x - 4} \right) + 3\left( {x - 4} \right)}}{{x\left( {x - 1} \right)}} = \frac{{\left( {x + 1} \right)\left( {x - 4} \right)}}{{x\left( {x + 1} \right)}}.\frac{{x\left( {x - 1} \right)}}{{\left( {x - 4} \right)\left( {x + 3} \right)}} = \frac{{x - 1}}{{x + 3}}\)