\(x^3+1=x\left(x+1\right)\)

\(\Leftrightarrow x^3+1-x\left(x+1\right)=0\)

\(\Leftrightarrow x^3+1-x^2-x=0\)

\(\Leftrightarrow x^3-x^2-x+1\)

\(\Leftrightarrow x^2.\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x^2=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=\pm1\end{cases}}}\)

Vậy x = {1;-1}

\(x^3+1=x\left(x+1\right)\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)=x\left(x+1\right)\)    
\(\Leftrightarrow x^2+2x+1=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\)
\(x=-1\)

Thử: 

Gọi 2/5 stn=1/6 sth = x

=> stn=5x/2 ; sth=6x

stn+sth=17x/2=51

x=51:17x2=6

Vậy stn=5x/2=15 ; sth=6x=36

Gọi st1 là x

Số t2 la : 51-x 

Theo đề bài ta có pt:   \(\frac{2}{5}x=\frac{1}{6}\left(51-x\right)\)

\(\Leftrightarrow\)                         \(\frac{2}{5}x=\frac{17}{2}-\frac{1}{6}x\)

\(\Leftrightarrow\)          \(\frac{2}{5}x+\frac{1}{6}x=\frac{17}{2}\)

\(\Leftrightarrow\)                      \(\frac{17}{30}x=\frac{17}{2}\)

\(\Leftrightarrow\)                              \(x=15\)

Vậy :...................................