Question

x+1/x^2+x+1-x-1/x^2-x+1=2(x+2)^2/x^6-1

Answer 1

\(\Leftrightarrow\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{2\left(x+2\right)^2}{\left(x+1\right)\left(x-1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}\)

Suy ra: \(\left(x+1\right)^2\cdot\left(x^2-x+1\right)-\left(x-1\right)^2\cdot\left(x^2+x+1\right)=2\left(x+2\right)^2\)

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

\(\Leftrightarrow x^4+x^3+x+1-x^4+x^3+x-1=2\left(x+2\right)^2\)

\(\Leftrightarrow2x^3+2x-2\left(x+2\right)^2=0\)

\(\Leftrightarrow2x^2\left(x+1\right)-2\left(x+2\right)^2=0\)