Question

[Mimi Queen Ni]

(x+1)/(x^2+x+1) - (x-1)/(x^2-x-1) = [2.(x+2)^2]/(x^6-1)

Answer 1

[Minh]

\(ĐKXĐ:x\ne\pm1\)

Ta có : \(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{x^6-1}\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}=\frac{2\left(x+2\right)^2}{\left(x^3+1\right)\left(x^3-1\right)}\)

\(\Leftrightarrow\frac{x^3+1-x^3+1}{\left(x^2+x+1\right)\left(x^2-x+1\right)}-\frac{2\left(x+2\right)^2}{\left(x+1\right)\left(x^2-x+1\right)\left(x-1\right)\left(x^2+x+1\right)}=0\)

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

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

\(\Leftrightarrow2\left(x^2-1\right)-2\left(x^2+4x+4\right)=0\)

\(\Leftrightarrow2x^2-2-2x^2-8x-8=0\)

\(\Leftrightarrow-8x-10=0\)

\(\Leftrightarrow x=-\frac{5}{4}\)

Vậy \(x=-\frac{5}{4}\) là nghiệm của phương trình.