Thực hiện các phép tính sau:

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

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

Chú ý: Đề câu b) trong sách chưa chính xác, cần sửa thành \(\frac{{{x^2} - 3x - 4}}{{{x^2} + x}}:\frac{{{x^2} - x - 12}}{{{x^2} - x}}\).

Lời giải

a) \[\frac{{2x + 4}}{{x + 3}} + \frac{3}{x} - \frac{6}{{{x^2} + 3x}}\]

\[ = \frac{{x\left( {2x + 4} \right)}}{{x\left( {x + 3} \right)}} + \frac{{3\left( {x + 3} \right)}}{{x\left( {x + 3} \right)}} - \frac{6}{{x\left( {x + 3} \right)}}\]

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

\[ = \frac{{2{x^2} + 7x\, + 3}}{{x(x + 3)}}\]

\[ = \frac{{\left( {2{x^2} + 6x} \right) + \left( {x\, + 3} \right)}}{{x(x + 3)}}\]

\[ = \frac{{2x(x + 3) + (x + 3)}}{{x(x + 3)}}\]

\[ = \frac{{(x + 3)(2x + 1)}}{{x(x + 3)}}\]

\[ = \frac{{2x + 1}}{x}\].

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

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

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

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

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

\( = \frac{{x - 1}}{{x + 3}}\).