Пожалуйста!!!!! Помогите!Алгебра x/(x²-3) + (x-8)/x – (72+3x-7x²)/(x³-9)= 8/(10-2q) +

Ответы 1

2) $\frac{8}{10-2q} + \frac{5-q}{q^2+5q} + \frac{q^3+5q^2-15q+25}{q^3-25q}= \\ \\ = \frac{8}{2(5-q)} + \frac{5-q}{q(q+5)} + \frac{q^3+5q^2-15q+25}{(q^2-25)q} =$ $=- \frac{8}{2(q-5)} + \frac{5-q}{q(q+5)} + \frac{q^3+5q^2-15q+25}{(q-5)(q+5)q}= \\ \\ = \frac{-8q(q+5)+(5-q)\cdot2(q-5)+2(q^3+5q^2-15q+25)}{2(q-5)(q+5)q}= \\ \\ =\frac{-8q^2-40q-2q^2+50+2q^3+10q^2-30q+50}{2(q-5)(q+5)q}= \\ \\ =\frac{2q^3-70q+100}{2(q-5)(q+5)q}= \\ \\= \frac{q^3-35q+50}{q(q-5)(q+5)}= \\ \\=\frac{(q-5)(q^2+5q-10)}{q(q-5)(q+5)}= \\ \\=\frac{q^2+5q-10}{q(q+5)}$ 3) $\frac{4-m}{m^2-4m} - \frac{m-4}{12+3m} - \frac{(48-8m+11m^2)}{48m-3m^3} =$ $=\frac{4-m}{m(m-4)} - \frac{m-4}{3(m+4)} - \frac{(48-8m+11m^2)}{3m(16-m^2)} =$ $=\frac{3(4-m)(m+4)}{3m(m-4)(m+4)} - \frac{3m(m-4)(m-4)}{3m(m+4)(m-4)} +\frac{(48-8m+11m^2)}{3m(m-4)(m+4)} =$ $=\frac{3(4-m)(m+4)-3m(m-4)^2+(48-8m+11m^2)}{3m(m-4)(m+4)} =$ $=\frac{48-3m^2-3m(m-4)^2+48-8m+11m^2}{3m(m-4)(m+4)} =$ $\frac{96-8m+8m^2-3m(m-4)^2}{3m(m-4)(m+4)} =\frac{8(m^2-m+12)-3m(m-4)^2}{3m(m-4)(m+4)} =$ $\frac{8(m-4)(m+3)-3m(m-4)^2}{3m(m-4)(m+4)} =\frac{(m-4)(8m+24-3m(m-4))}{3m(m-4)(m+4)} =\frac{8m+24-3m^2+12m}{3m(m+4)} =$ $\frac{-3m^2+20m+24}{3m(m+4)}$
- Автор:
  beckham
- 6 лет назад
- 0
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