Giải bài 4 trang 16 vở thực hành Toán 8 tập 2

Đề bài

Thực hiện các phép tính sau:

a) \(\frac{{{x^2} + 4{\rm{x}} + 4}}{{{x^2} - 4}} + \frac{x}{{2 - x}} + \frac{{4 - x}}{{5{\rm{x}} - 10}}\)

b) \(\frac{x}{{{x^2} + 1}} - \left( {\frac{3}{{x + 6}} + \frac{{x - 2}}{{x + 4}}} \right) + \left[ {\frac{3}{{x + 6}} - \left( {\frac{1}{{{x^2} + 1}} - \frac{{x - 2}}{{x + 4}}} \right)} \right]\)

Lời giải chi tiết

a)

\(\begin{array}{*{20}{l}}{\frac{{{x^2} + 4{\rm{x}} + 4}}{{{x^2} - 4}} + \frac{x}{{2 - x}} + \frac{{4 - x}}{{5{\rm{x}} - 10}}}\\{ = \frac{{{{\left( {x + 2} \right)}^2}}}{{\left( {x + 2} \right)\left( {x - 2} \right)}} + \frac{{ - x}}{{x - 2}} + \frac{{4 - x}}{{5\left( {x - 2} \right)}}}\\{ = \frac{{x + 2}}{{x - 2}} + \frac{{ - x}}{{x - 2}} + \frac{{4 - x}}{{5\left( {x - 2} \right)}}}\\{ = \frac{2}{{x - 2}} + \frac{{4 - x}}{{5(x - 2)}} = \frac{{ - x + 14}}{{5\left( {x - 2} \right)}}}\end{array}\)

b)

\(\begin{array}{*{20}{l}}{\frac{x}{{{x^2} + 1}} - \left( {\frac{3}{{x + 6}} + \frac{{x - 2}}{{x + 4}}} \right) + \left[ {\frac{3}{{x + 6}} - \left( {\frac{1}{{{x^2} + 1}} - \frac{{x - 2}}{{x + 4}}} \right)} \right]}\\{ = \frac{x}{{{x^2} + 1}} - \frac{3}{{x + 6}} - \frac{{x - 2}}{{x + 4}} + \frac{3}{{x + 6}} - \frac{1}{{{x^2} + 1}} + \frac{{x - 2}}{{x + 4}}}\\\begin{array}{l} = \left( {\frac{x}{{{x^2} + 1}} - \frac{1}{{{x^2} + 1}}} \right) + \left( {\frac{3}{{x + 6}} - \frac{3}{{x + 6}}} \right) + \left( {\frac{{x - 2}}{{x + 4}} - \frac{{x - 2}}{{x + 4}}} \right)\\ = \frac{{x - 1}}{{{x^2} + 1}} + 0 + 0 = \frac{{x - 1}}{{{x^2} + 1}}\end{array}\end{array}\)