Giải bài 3 trang 42 SGK Toán 8 tập 1 - Cánh diều

Đề bài

Thực hiện phép tính:

a) \(\frac{1}{{x - 2}} - \frac{1}{{x + 1}}\)

b) \(\frac{{12}}{{{x^2} - 9}} - \frac{2}{{x - 3}}\)

c) \(\frac{1}{{xy - {x^2}}} - \frac{1}{{{y^2} - xy}}\)

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

Lời giải chi tiết

a)

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

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

b)

\(\begin{array}{l}\frac{{12}}{{{x^2} - 9}} - \frac{2}{{x - 3}} = \frac{{12}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} - \frac{2}{{x - 3}}\\ = \frac{{12}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} - \frac{{2\left( {x + 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \frac{{12 - 2{\rm{x}} - 6}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\ = \frac{{6 - 2{\rm{x}}}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \frac{{ - 2\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \frac{{ - 2}}{{x + 3}}\end{array}\)

c)

\(\begin{array}{l}\frac{1}{{xy - {x^2}}} - \frac{1}{{{y^2} - xy}} = \frac{1}{{x\left( {y - x} \right)}} - \frac{1}{{y\left( {y - x} \right)}}\\ = \frac{y}{{xy\left( {y - x} \right)}} - \frac{x}{{xy\left( {y - x} \right)}} = \frac{{y - x}}{{xy\left( {y - x} \right)}} = \frac{1}{{xy}}\end{array}\)

d)

 \(\begin{array}{l}\frac{{2{\rm{x}}}}{{{x^2} - 1}} - \frac{3}{{2 + 2{\rm{x}}}} + \frac{1}{{2 - 2{\rm{x}}}}\\ = \frac{{2{\rm{x}}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} - \frac{3}{{2\left( {x + 1} \right)}} - \frac{1}{{2{\rm{x}} - 2}}\\ = \frac{{2{\rm{x}}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} - \frac{{3\left( {x - 1} \right)}}{{2\left( {x + 1} \right)\left( {x - 1} \right)}} - \frac{1}{{2\left( {x - 1} \right)}}\\ = \frac{{{\rm{4x}}}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}} - \frac{{3\left( {x - 1} \right)}}{{2\left( {x + 1} \right)\left( {x - 1} \right)}} - \frac{{1\left( {x + 1} \right)}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\ = \frac{{{\rm{4x}} - 3{\rm{x}} + 3 - x - 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{2}{{2\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\end{array}\)