If the sum of the coefficients in the expansion of (x + ax) (2x − 1x) 5 is 2, then the constant term in the expansion is () A. -40B. -20C. 20D. 40

If the sum of the coefficients in the expansion of (x + ax) (2x − 1x) 5 is 2, then the constant term in the expansion is () A. -40B. -20C. 20D. 40

Let's get the coefficients of the expansion, and get the coefficients of the expansion and the coefficients of the expansion and the coefficients of the expansion, and the coefficients of the expansion and the coefficients of the expansion and the coefficients and the coefficients of the expansion, and the coefficients of the expansion and the coefficients of the expansion are all the coefficients and the coefficients of all the coefficients of all the coefficients of the expansion and the coefficients of the expansion are all the coefficients of the expansion, and the coefficients of the expansion are all the coefficients of the expansion and the coefficients of the expansion are all the coefficients of the expansion and the coefficients of the expansion and the coefficients of the expansion \\\\\\\\\\\\\\\\\\\\\\\\\\4c53 = 40, so D is selected

If the sum of the coefficients in the expansion of (x + ax) (2x − 1x) 5 is 2, then the constant term in the expansion is ()
A. -40B. -20C. 20D. 40

Let X in the binomial be 1, and the sum of coefficients of the expansion is 1 + a ∪ 1 + a = 2 ∪ a = 1 ∪ (x + ax) (2x − LX) 5 = (x + 1x) (2x − LX) 5 = x (2x − LX) 5 + 1x (2x − LX) 5 ∪ in the expansion, the coefficients of LX and X with constant term of (2x − LX) 5 and the general term of ∫ (2x − LX) 5 are tr + 1 = (- 1) r25-rc5rx5 -

(2x + 1) ^ 2 - (x-3) (2x-1) = 3x, what is the coefficient of quadratic term, coefficient of primary term and constant term

（2x+1)^2-(x-3)(2x-1)=3x
4x^2+4x+1-2x^2+7x-3-3x=0
2x^2+8x-2=0
x^2+4x-1=0
The coefficient of quadratic term is 1, the coefficient of primary term is 4, and the constant term is - 1