\(f\left(x+3\right)-3f\left(x+2\right)+3f\left(x+1\right)\)

\(=a\left(x+3\right)^2+b\left(x+3\right)+c-3\left[a\left(x+2\right)^2+b\left(x+2\right)+c\right]+3\left[a\left(x+1\right)^2+b\left(x+1\right)+c\right]\)

\(=3a\left(x+1\right)^2+a\left(x+3\right)^2-3a\left(x+2\right)^2+bx+c\)

\(=ax^2+bx+c\)

Ta có:

f(x+3) = a(x+3)²+ b(x+3) +c=ax²+ (6a+b) x+ 9a+ 3b+c

f(x+2) = a(x+2)²+ b(x+2) +c=ax²+ (4a+b) x+ 4a+ 2b+c

f (x+1) = a(x+1)²+ b(x+1) +c=ax²+ (2a+b) x+ 2a+ 2b+c

Suy ra: (x+ 3) -3f( x+ 2) +3f( x+ 1)= ax²+ bx+ c

Chọn D.

Đáp án D

Ta có: (x-2)⁵=(x-2)³.(x-2)²=(x³-6x²+12x-8)(x²-4x+4)=x⁵-6x⁴+12x³-8x²-4x⁴+24x³-48x²+32x+4x³-24x²+48x-32 = x⁵-10x⁴+40x³-32x²+80x-32

(x-1)⁴=(x-1)²(x-1)² = (x²-2x+1)(x²-2x+1)=x⁴-2x³+x²-2x³+4x²-2x+x²-2x+1=x⁴-4x³+6x²-4x+1

Và: (x+1)²=x²+2x+1

=> P(x)= (x⁵-10x⁴+40x³-32x²+80x-32) + (x⁴-4x³+6x²-4x+1) + x³ +(x²+2x+1)+x+2

=> P(x)= x⁵-10x⁴+40x³-32x²+80x-32 + x⁴-4x³+6x²-4x+1 + x³ +x²+2x+1+x+2

=> P(x)= x⁵-9x⁴+37x³-25x²+79x-28

=> a=1; b=-9; c=37; d=-25; e=79; f=-28

=> a+3b+c+3d+e+3f = 1+3(-9)+37+3(-25)+79+3(-28) = 1-27+37-75+79-84=(1+37+79)-(27+75+84)=117-186

=> a+3b+c+3d+e+3f = - 69