Đề phải là \(A=1-\left(\dfrac{2x^2+x-1}{1-x^2}+\dfrac{2x^3-x+x^2}{1+x^3}\right).\dfrac{\left(1-x\right)\left(x^2-x\right)}{2x-1}\) thì mới làm đc.
Điều kiện xác định: \(\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne\dfrac{1}{2}\end{matrix}\right.\)
\(A=1-\left[\dfrac{2x^2+x-1}{\left(1-x\right)\left(1+x\right)}+\dfrac{x\left(2x^2+x-1\right)}{\left(1+x\right)\left(1-x+x^2\right)}\right].\dfrac{\left(1-x\right).x\left(x-1\right)}{2x-1}\)
\(=1-\left\{\dfrac{\left(2x^2-x\right)+\left(2x-1\right)}{\left(1-x\right)\left(1+x\right)}+\dfrac{x\left[\left(2x^2-x\right)+\left(2x-1\right)\right]}{\left(1+x\right)\left(1-x+x^2\right)}\right\}.\dfrac{\left(1-x\right).x\left(x-1\right)}{2x-1}\)
\(=1-\left\{\dfrac{x\left(2x-1\right)+\left(2x-1\right)}{\left(1-x\right)\left(1+x\right)}+\dfrac{x\left[x\left(2x-1\right)+\left(2x-1\right)\right]}{\left(1+x\right)\left(1-x+x^2\right)}\right\}.\dfrac{\left(1-x\right).x\left(x-1\right)}{2x-1}\)
\(=1-\left[\dfrac{\left(x+1\right)\left(2x-1\right)}{\left(1-x\right)\left(x+1\right)}+\dfrac{x\left(x+1\right)\left(2x-1\right)}{\left(x+1\right)\left(1-x+x^2\right)}\right].\dfrac{\left(1-x\right).x\left(x-1\right)}{2x-1}\)
\(=1-\left[\dfrac{2x-1}{1-x}+\dfrac{x\left(2x-1\right)}{1-x+x^2}\right].\dfrac{\left(1-x\right).x\left(x-1\right)}{2x-1}\)
\(=1-\dfrac{\left(2x-1\right)\left(1-x+x^2\right)+x\left(2x-1\right)\left(1-x\right)}{\left(1-x\right)\left(1-x+x^2\right)}.\dfrac{\left(1-x\right).x\left(x-1\right)}{2x-1}\)
\(=1-\dfrac{\left(2x-1\right)\left[\left(1-x+x^2\right)+x\left(1-x\right)\right]}{\left(1-x\right)\left(1-x+x^2\right)}.\dfrac{\left(1-x\right).x\left(x-1\right)}{2x-1}\)
\(=1-\dfrac{\left(2x-1\right)\left[1-x+x^2+x-x^2\right]}{\left(1-x\right)\left(1-x+x^2\right)}.\dfrac{\left(1-x\right).x\left(x-1\right)}{2x-1}\)
\(=1-\dfrac{\left(2x-1\right)}{\left(1-x\right)\left(1-x+x^2\right)}.\dfrac{\left(1-x\right).x\left(x-1\right)}{2x-1}\)
\(=1-\dfrac{x\left(2x-1\right)\left(1-x\right)\left(x-1\right)}{\left(2x-1\right)\left(1-x\right)\left(1-x+x^2\right)}\)
\(=1-\dfrac{x\left(x-1\right)}{1-x+x^2}\)
\(=\dfrac{1-x+x^2-x\left(x-1\right)}{1-x+x^2}\)
\(=\dfrac{1-x+x^2-x^2+x}{1-x+x^2}\)
\(=\dfrac{1}{1-x+x^2}\)
