1. Đ
\(\dfrac{\left(x-8\right)^3}{2\left(8-x\right)}=\dfrac{\left(8-x\right)^3:\left(8-x\right)}{2\left(8-x\right):\left(8-x\right)}=\dfrac{\left(8-x\right)^2}{2}\)
2. ta có:\(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)
MTC: \(x^3-1\)
NTP:
NTP:\(x-1\)
\(\dfrac{3x}{x^3-1}\) giữ nguyên
\(\dfrac{x-1}{x^2+x+1}=\dfrac{\left(x-1\right)^2}{x^3-1}\) ta nhân cho x-1
bây giờ mới xong đó