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)

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