ĐKXĐ: x khác -1
Bạn chọn MTC là x^3+1 nhé
a) Rút gọn đc :
a, ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ne0\\x^2-x+1\ne0\\x+2\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne-1\\x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\ne0\\x\ne-2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x\ne-1\\\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ne0\\x\ne-2\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne-1\\x\ne-2\end{matrix}\right.\)
Ta có : \(Q=\left(\frac{1}{x+1}-\frac{6x-3}{x^3+1}+\frac{2}{x^2-x+1}\right):\left(x+2\right)\)
\(Q=\left(\frac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{6x-3}{x^3+1}+\frac{2\left(x+1\right)}{\left(x^2-x+1\right)\left(x+1\right)}\right):\left(x+2\right)\)
\(Q=\left(\frac{x^2-x+1-6x+3+2\left(x+1\right)}{x^3+1}\right):\left(x+2\right)\)
\(Q=\left(\frac{x^2-x+1-6x-3+2x+2}{x^3+1}\right).\frac{1}{x+2}\)
\(Q=\left(\frac{x^2-5x}{x^3+1}\right).\frac{1}{x+2}\) = \(\frac{x^2-5x}{x^4+2x^3+x+2}\)
a, rút gọn
b,tìm giá trị lớn nhất
