Đề phải là \(H=\left(\dfrac{1}{x+1}-\dfrac{3}{x^3+1}+\dfrac{3}{x^2-x+1}\right).\dfrac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\dfrac{2x-2}{x^2+2x}\) thì mới làm đc.
Điều kiện xác định: \(\left\{{}\begin{matrix}x\ne-2\\x\ne-1\\x\ne0\end{matrix}\right.\)
\(H=\left[\dfrac{1}{x+1}-\dfrac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{3}{x^2-x+1}\right].\dfrac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\dfrac{2x-2}{x\left(x+2\right)}\)
\(=\dfrac{x^2-x+1-3+3\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\dfrac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\dfrac{2x-2}{x\left(x+2\right)}\)
\(=\dfrac{x^2-x+1-3+3x+3}{\left(x+1\right)\left(x^2-x+1\right)}.\dfrac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\dfrac{2x-2}{x\left(x+2\right)}\)
\(=\dfrac{x^2+2x+1}{\left(x+1\right)\left(x^2-x+1\right)}.\dfrac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\dfrac{2x-2}{x\left(x+2\right)}\)
\(=\dfrac{\left(x+1\right)^2}{\left(x+1\right)\left(x^2-x+1\right)}.\dfrac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\dfrac{2x-2}{x\left(x+2\right)}\)
\(=\dfrac{3\left(x+1\right)^2.\left(x^2-x+1\right)}{\left(x+1\right)^2.\left(x^2-x+1\right)\left(x+2\right)}-\dfrac{2x-2}{x\left(x+2\right)}\)
\(=\dfrac{3}{x+2}-\dfrac{2x-2}{x\left(x+2\right)}\)
\(=\dfrac{3x-\left(2x-2\right)}{x\left(x+2\right)}\)
\(=\dfrac{3x-2x+2}{x\left(x+2\right)}\)
\(=\dfrac{x+2}{x\left(x+2\right)}\)
\(=\dfrac{1}{x}\)
