Упростите выражение: (1/2-8p + 1/64p³-1) : 1+4p/1+4p+16p²

Упростим выражение:

$$\left(\frac{1}{2-8p} + \frac{1}{64p^3-1}\right) : \frac{1+4p}{1+4p+16p^2}$$

Разложим знаменатели на множители:

$$2-8p = 2(1-4p)$$ $$64p^3-1 = (4p-1)(16p^2+4p+1)$$

Тогда выражение примет вид:

$$\left(\frac{1}{2(1-4p)} + \frac{1}{(4p-1)(16p^2+4p+1)}\right) : \frac{1+4p}{1+4p+16p^2} = $$ $$\left(\frac{-1}{2(4p-1)} + \frac{1}{(4p-1)(16p^2+4p+1)}\right) : \frac{1+4p}{1+4p+16p^2} = $$ $$\frac{-1(16p^2+4p+1) + 2}{2(4p-1)(16p^2+4p+1)} : \frac{1+4p}{1+4p+16p^2} = $$ $$\frac{-16p^2-4p-1 + 2}{2(4p-1)(16p^2+4p+1)} : \frac{1+4p}{1+4p+16p^2} = $$ $$\frac{-16p^2-4p+1}{2(4p-1)(16p^2+4p+1)} : \frac{1+4p}{16p^2+4p+1} = $$ $$\frac{-(16p^2+4p-1)}{2(4p-1)(16p^2+4p+1)} \cdot \frac{16p^2+4p+1}{4p+1} = $$ $$\frac{-(16p^2+4p-1)(16p^2+4p+1)}{2(4p-1)(16p^2+4p+1)(4p+1)} = $$ $$\frac{-(16p^2+4p-1)}{2(4p-1)(4p+1)} = \frac{-(16p^2+4p-1)}{2(16p^2-1)} = \frac{-16p^2-4p+1}{32p^2-2}$$

Ответ: $$ \frac{-16p^2-4p+1}{32p^2-2}$$