ĐKXĐ : \(\left\{{}\begin{matrix}x^2-x+1\ge0\\x+1>0\\x^3+1>0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x^2-2x\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\ge0\\x+1>0\\x^3+1>0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge0\\x+1>0\\x^3+1>0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x>-1\\x>-1\end{matrix}\right.\)
=> \(x>-1.\)
Ta có : \(5\sqrt{\frac{x^2-x+1}{x+1}}-\left(x+1\right)\sqrt{\frac{1}{x^3+1}}=4\)
=> \(5\sqrt{\frac{x^2-x+1}{x+1}}-\sqrt{\frac{\left(x+1\right)^2}{\left(x+1\right)\left(x^2-x+1\right)}}=4\)
=> \(5\sqrt{\frac{x^2-x+1}{x+1}}-\sqrt{\frac{x+1}{x^2-x+1}}=4\)
Đặt \(\sqrt{\frac{x^2-x+1}{x+1}}=a\) => \(\frac{1}{a}=\sqrt{\frac{x+1}{x^2-x+1}}\) ( ĐK : \(a\ge0\) ) ta được phương trình :
\(5a-\frac{1}{a}=4\)
=> \(\frac{5a^2}{a}-\frac{1}{a}=4\)
=> \(5a^2-1=4a\)
=> \(5a^2-1-5a+a=0\)
=> \(5a\left(a-1\right)+\left(a-1\right)=0\)
=> \(\left(5a+1\right)\left(a-1\right)=0\)
=> \(\left[{}\begin{matrix}5a+1=0\\a-1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}a=-\frac{1}{5}\left(L\right)\\a=1\left(TM\right)\end{matrix}\right.\)
=> \(a=1\)
- Thay \(\sqrt{\frac{x^2-x+1}{x+1}}=a\) vào phương trình trên ta được :
\(\sqrt{\frac{x^2-x+1}{x+1}}=1\)
=> \(\frac{x^2-x+1}{x+1}=1\)
=> \(x^2-x+1=x+1\)
=> \(x^2-x+1-x-1=0\)
=> \(x^2=2x\)
=> \(x=2\) ( TM )
Vậy phương trình có nghiệm là x = 2 .
