Question

[Bảo Vi]

Với m + t = 1. Chứng minh rằng với \(m\ne0,t\ne0\)ta có: \(\frac{t}{m^3-1}-\frac{m}{t^3-1}=\frac{2\left(m-t\right)}{m^2t^2+3}\)

Answer 1

[Đinh Đức Hùng]

\(VT=\frac{t}{\left(m-1\right)\left(m^2+m+1\right)}-\frac{m}{\left(t-1\right)\left(t^2+t+1\right)}=\frac{t}{-t\left(m^2+m+1\right)}-\frac{m}{-m\left(t^2+t+1\right)}\)\(=\frac{-1}{m^2+m+1}+\frac{1}{t^2+t+1}=\frac{-t^2-t-1+m^2+m+1}{\left(m^2+m+1\right)\left(t^2+t+1\right)}\)

\(=\frac{\left(m-t\right)\left(m+t\right)+m-t}{m^2t^2+mt\left(m+t\right)+m^2+t^2+mt+\left(m+t\right)+1}\)

\(=\frac{2\left(m-t\right)}{m^2t^2+\left(m^2+t^2+2mt\right)+2}=\frac{2\left(m-t\right)}{m^2t^2+\left(m+t\right)^2+2}=\frac{2\left(m-t\right)}{m^2t^2+3}=VP\)