ĐKXĐ : \(\left\{{}\begin{matrix}x^2+x+1\ne0\\x^2-x+1\ne0\\x^6-1\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ne0\\\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ne0\\x^6\ne1\end{matrix}\right.\)
=> \(x\ne1\)
Ta có : \(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{x^6-1}\)
=> \(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{\left(x^3+1\right)\left(x^3-1\right)}\)
=> \(\frac{\left(x^2-1\right)\left(x^3+1\right)}{\left(x^3-1\right)\left(x^3+1\right)}-\frac{\left(x^2-1\right)\left(x^3-1\right)}{\left(x^3+1\right)\left(x^3-1\right)}=\frac{2\left(x+2\right)^2}{\left(x^3+1\right)\left(x^3-1\right)}\)
=> \(\left(x^2-1\right)\left(x^3+1\right)-\left(x^2-1\right)\left(x^3-1\right)=2\left(x+2\right)^2\)
=> \(\left(x^2-1\right)\left(x^3+1-x^3+1\right)=2\left(x+2\right)^2\)
=> \(2\left(x^2-1\right)=2\left(x+2\right)^2\)
=> \(x^2-1=\left(x+2\right)^2\)
=> \(x^2-1=x^2+4x+4\)
=> \(x^2-1-x^2-4x-4=0\)
=> \(x=-\frac{5}{4}\) ( TM )
Vậy phương trình trên có tập nghiệm là \(S=\left\{-\frac{5}{4}\right\}\)
1: Ta có: \(\frac{1}{3-x}-\frac{1}{x+1}=\frac{x}{x-3}-\frac{\left(x-1\right)^2}{x^2-2x-3}\)
⇔\(\frac{1}{3-x}-\frac{1}{x+1}-\frac{x}{x-3}+\frac{\left(x-1\right)^2}{x^2-2x-3}=0\)
⇔\(\frac{1}{3-x}+\frac{x}{3-x}-\frac{1}{x+1}+\frac{\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}=0\)
⇔\(\frac{1}{3-x}+\frac{x}{3-x}-\frac{1}{x+1}-\frac{\left(x-1\right)^2}{\left(3-x\right)\left(x+1\right)}=0\)
⇔\(\frac{\left(1+x\right)\left(x+1\right)}{\left(3-x\right)\left(x+1\right)}-\frac{3-x}{\left(3-x\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(3-x\right)\left(x+1\right)}=0\)
⇔\(x^2+2x+1-3+x-\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2+2x+1-3+x-x^2+2x-1=0\)
\(\Leftrightarrow5x-3=0\)
\(\Leftrightarrow5x=3\)
hay \(x=\frac{3}{5}\)
Vậy: \(x=\frac{3}{5}\)
Dễ mà bạn
