Question

Rút gọn:

(2x-1)(4x^2-3x+1)+(2x+1)(4x^2+3x+1)

Answer 1

ta có : \(\left(2x-1\right)\left(4x^2-3x+1\right)+\left(2x+1\right)\left(4x^2+3x+1\right)\)

\(=2x\left(4x^2-3x+1\right)-\left(4x^2-3x+1\right)+2x\left(x^2+3x+1\right)+\left(4x^2+3x+1\right)\)

\(=2x\left(4x^2-3x+1\right)+2x\left(4x^2+3x+1\right)-\left(4x^2-3x+1\right)+\left(4x^2+3x+1\right)\)

\(=2x\left(4x^2-3x+1+4x^2+3x+1\right)-\left(4x^2-3x+1-4x^2-3x-1\right)\)

\(=2x\left(8x^2+2\right)-\left(-6x\right)=16x^3+4x+6x=16x^3+10x\)