Làm:
\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\)
(=) \(\frac{\left(x+1\right)x\left(x^2-x+1\right)}{x\left(x^4+x^2+1\right)}-\frac{x\left(x-1\right)\left(x^2+x+1\right)}{x\left(x^2-x+1\right)}=\frac{3}{x\left(x^4+x^2+1\right)}\)
=> x(x3+1)- x(x3-1)=3
<=> x4+x-x4+x = 3
<=> 2x=3
<=> x =\(\frac{3}{2}\)
Kl: Vậy nghiệm của pt là x=\(\frac{3}{2}\)
\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\) \(\left(ĐKXĐ:x\ne0\right)\)
\(\Leftrightarrow\frac{x\left(x^2-x+1\right)\left(x+1\right)}{x\left(x^4+x^2+1\right)}-\frac{x\left(x^2+x+1\right)\left(x-1\right)}{x\left(x^4+x^2+1\right)}=\frac{3}{x\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow x^4+x-x^4+x=3\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Vậy phương trình có nghiệm \(x=\frac{3}{2}.\)
Ta có: \(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}-\frac{3}{x\left(x^4+x^2+1\right)}=0\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x^2-x+1\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}-\frac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2-x+1\right)\left(x^2+x+1\right)}-\frac{3}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)-3=0\)
\(\Leftrightarrow x^3+1-\left(x^3-1\right)-3=0\)
\(\Leftrightarrow x^3+1-x^3+1-3=0\)
\(\Leftrightarrow-1\ne0\)
Vậy: x∈∅
