先因式分解,再求值。(x+1)的平方(2x-3)+(x+1)(2x_3)的平方-(X+1)(3-2x),其中X=2
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(x+1)^2(2x-3)+(x+1)(2x-3)^2-(x+1)(3-2x)
=(x+1)^2(2x-3)+(x+1)(2x-3)^2+(x+1)(2x-3)
=(2x-3)[(x+1)^2+(x+1)(2x-3)+(x+1)]
=(2*2-3)[(x+1)^2+(x+1)(2x-3)+(x+1)]
=(x+1)^2+(x+1)(2x-3)+(x+1)
=(x+1)[(x+1)+(2x-3)+1]
=(x+1)(3x-1)
=(2+1)(3*2-1)
=3*5
=15
=(x+1)^2(2x-3)+(x+1)(2x-3)^2+(x+1)(2x-3)
=(2x-3)[(x+1)^2+(x+1)(2x-3)+(x+1)]
=(2*2-3)[(x+1)^2+(x+1)(2x-3)+(x+1)]
=(x+1)^2+(x+1)(2x-3)+(x+1)
=(x+1)[(x+1)+(2x-3)+1]
=(x+1)(3x-1)
=(2+1)(3*2-1)
=3*5
=15
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