b)
ĐKXĐ: \(x\ne-1\)
Ta có: \(\dfrac{5}{x^2-x+1}-\dfrac{1}{x+1}=\dfrac{4-7x^2}{1+x^3}\)
\(\Leftrightarrow\dfrac{5\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{4-7x^2}{\left(x+1\right)\left(x^2-x+1\right)}\)
Suy ra: \(5x+5-x^2+x-1-4+7x^2=0\)
\(\Leftrightarrow6x^2+6x=0\)
\(\Leftrightarrow6x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)
Vậy: S={0}
\(ĐKXĐ:x\ne-1\)
\(\Leftrightarrow\dfrac{5\left(x+1\right)}{\left(x^2-x+1\right)\left(x+1\right)}-\dfrac{x^2-x+1}{\left(x^2-x+1\right)\left(x+1\right)}=\dfrac{4-7x^2}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\Leftrightarrow\dfrac{5x+5-x^2+x-1}{\left(x^2-x+1\right)\left(x+1\right)}=\dfrac{4-7x^2}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\Rightarrow6x+4=4-7x^2\Leftrightarrow7x^2+6x=0\Leftrightarrow x\left(7x+6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-\dfrac{6}{7}\left(TM\right)\end{matrix}\right.\)
\(\dfrac{5}{x^2-x+1}-\dfrac{1}{x+1}=\dfrac{4-7x^2}{1+x^3}\) ĐKXĐ: \(x\ne-1\)
⇔\(5\left(x+1\right)-1\left(x^2-x+1\right)-\left(4-7x^2\right)=0\)
⇔\(5x+5-x^2+x-1-4+7x^2=0\)
⇔\(6x^2+6x=0\)
⇔\(6x\left(x+1\right)=0\)
⇔\(\left\{{}\begin{matrix}6x=0\\x+1=0\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x=0\\x=-1\left(\notinĐKXĐ\right)\end{matrix}\right.\)
Vậy \(x=0\)
