Giải bài 2.14 trang 46 SGK Toán 8 – Cùng khám phá

Đề bài

Thực hiện các phép tính sau:

a) \(\frac{4}{{y – 5}} + \frac{2}{{2y + 1}}\)

b) \(\frac{{6x}}{{3x – 2}} – \frac{{x – 10}}{{2 – 3x}}\)

c)\(\frac{b}{{2{a^2} – ab}} + \frac{{4a}}{{{b^2} – 2ab}}\)

d)\(\frac{m}{{{{\left( {m – n} \right)}^2}}} – \frac{{{m^2}}}{{{n^2} – {m^2}}}\).

Lời giải chi tiết

a) \(\frac{4}{{y – 5}} + \frac{2}{{2y + 1}} = \frac{{8y + 4}}{{\left( {y + 5} \right)\left( {2y + 1} \right)}} + \frac{{2y – 10}}{{\left( {y + 5} \right)\left( {2y + 1} \right)}} = \frac{{10y – 6}}{{\left( {y + 5} \right)\left( {2y + 1} \right)}}\)

b) \(\frac{{6x}}{{3x – 2}} – \frac{{x – 10}}{{2 – 3x}} = \frac{{6x}}{{3x – 2}} + \frac{{x – 10}}{{3x – 2}} = \frac{{7x – 10}}{{3x – 2}}\)

c) \(\frac{b}{{2{a^2} – ab}} + \frac{{4a}}{{{b^2} – 2ab}} = \frac{b}{{a\left( {2a – b} \right)}} + \frac{{4a}}{{b\left( {b – 2a} \right)}} = \frac{{ – {b^2}}}{{ab\left( {b – 2a} \right)}} + \frac{{4{a^2}}}{{ab\left( {b – 2a} \right)}} = \frac{{4{a^2} – b}}{{ab\left( {b – 2a} \right)}} = \frac{{ – 2a – b}}{{ab}}\)

d) \(\frac{m}{{{{\left( {m – n} \right)}^2}}} – \frac{{{m^2}}}{{{n^2} – {m^2}}} = \frac{m}{{{{\left( {m – n} \right)}^2}}} + \frac{{{m^2}}}{{{m^2} – {n^2}}} = \frac{{m\left( {m + n} \right) + {m^2}\left( {m – n} \right)}}{{{{\left( {m – n} \right)}^2}\left( {m + n} \right)}} = \frac{{{m^3} + {m^2} + mn – {m^2}n}}{{{{\left( {m – n} \right)}^2}\left( {m + n} \right)}}\)