Question

Giải phương trình: \(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\)

Answer 1

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=\(\frac{3}{2}\)

Answer 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 \(x=\frac{3}{2}.\)

Answer 3

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∈∅