Khi thực hiện các phép tính \(\dfrac{2x+1}{2x^2-x}+\dfrac{32x^2}{1-4x^2}+\dfrac{1-2x}{2x^2+x}\), ta có kết quả là
\(-8\).\(\dfrac{8\left(2x+1\right)}{2x-1}\).\(8\).\(\dfrac{8\left(2x-1\right)}{2x+1}\).Hướng dẫn giải:\(\dfrac{2x+1}{2x^2-x}+\dfrac{32x^2}{1-4x^2}+\dfrac{1-2x}{2x^2+x}\)
\(=\dfrac{2x+1}{x\left(2x-1\right)}+\dfrac{32x^2}{\left(1-2x\right)\left(1+2x\right)}+\dfrac{1-2x}{x\left(2x+1\right)}\)
\(=\dfrac{2x+1}{x\left(2x-1\right)}+\dfrac{-32x^2}{\left(2x-1\right)\left(1+2x\right)}+\dfrac{1-2x}{x\left(2x+1\right)}\)
\(=\dfrac{\left(2x+1\right)\left(2x+1\right)-32x^2.x+\left(1-2x\right)\left(2x-1\right)}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{4x^2+4x+1-32x^3-\left(4x^2-4x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{8x-32x^3}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{8x\left(1-4x^2\right)}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{8x\left(1-2x\right)\left(1+2x\right)}{x\left(2x-1\right)\left(2x+1\right)}=-8\).
