Giải bài 2 trang 25 vở thực hành Toán 8 tập 2

Đề bài

Rút gọn biểu thức sau:

a) \(\frac{2}{{3{\rm{x}}}} + \frac{x}{{x - 1}} + \frac{{6{{\rm{x}}^2} - 4}}{{2{\rm{x}}\left( {1 - x} \right)}}\)

b) \(\frac{{{x^3} + 1}}{{1 - {x^3}}} + \frac{x}{{x - 1}} - \frac{{x + 1}}{{{x^2} + x + 1}}\)

c) \(\left( {\frac{2}{{x + 2}} - \frac{2}{{1 - x}}} \right).\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\)

d) \(1 + \frac{{{x^3} - x}}{{{x^2} + 1}}\left( {\frac{1}{{1 - x}} - \frac{1}{{1 - {x^2}}}} \right)\)

Lời giải chi tiết

a) \(\frac{2}{{3{\rm{x}}}} + \frac{x}{{x - 1}} + \frac{{6{x^2} - 4}}{{2x\left( {1 - x} \right)}}\)\( = \frac{2}{{3{\rm{x}}}} + \frac{{ - x}}{{1 - x}} + \frac{{3{{\rm{x}}^2} - 2}}{{x\left( {1 - x} \right)}}\)\( = \frac{{2 - 2x - 3{x^2} + 9{x^2} - 6}}{{3x\left( {1 - x} \right)}}\)

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

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

c) Ta có: \(\frac{2}{{x + 2}} - \frac{2}{{1 - x}} = \frac{{2\left( {1 - x} \right) - 2\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {1 - x} \right)}} = \frac{{2 - 2{\rm{x}} - 2{\rm{x}} - 4}}{{\left( {x + 2} \right)\left( {1 - x} \right)}} = \frac{{ - 4x - 2}}{{\left( {x + 2} \right)\left( {1 - x} \right)}} = \frac{{2\left( {2x + 1} \right)}}{{\left( {x + 2} \right)\left( {x - 1} \right)}}\);

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

Do đó

\(\begin{array}{l}\left( {\frac{2}{{x + 2}} - \frac{2}{{1 - x}}} \right).\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}} = \frac{{2\left( {2x + 1} \right)}}{{\left( {x + 2} \right)\left( {x - 1} \right)}}.\frac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{\left( {2x - 1} \right)\left( {2x + 1} \right)}}\\ = \frac{{2(x - 2)}}{{\left( {2x - 1} \right)\left( {x - 1} \right)}}\end{array}\)

d) Ta có: \(\frac{1}{{1 - x}} - \frac{1}{{1 - {x^2}}} = \frac{{1 + x}}{{1 - {x^2}}} - \frac{1}{{1 - {x^2}}} = \frac{x}{{1 - {x^2}}} = \frac{x}{{\left( {1 - x} \right)\left( {1 + x} \right)}}\).

Do đó \(1 + \frac{{{x^3} - x}}{{{x^2} + 1}}\left( {\frac{1}{{1 - x}} - \frac{1}{{1 - {x^2}}}} \right) = 1 + \frac{{x\left( {x - 1} \right)\left( {x + 1} \right)}}{{{x^2} + 1}}.\frac{x}{{\left( {1 - x} \right)\left( {1 + x} \right)}}\)

\( = 1 + \frac{{ - {x^2}}}{{{x^2} + 1}}\)\( = \frac{{{x^2} + 1 - {x^2}}}{{{x^2} + 1}}\)\( = \frac{1}{{{x^2} + 1}}\).