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(D) To find $a$,the sum of all coefficients,we set $x=1$ in the expression: $a = (1-2(1)+2(1)^2)^{2023} 	imes (3-4(1)^2+2(1)^3)^{2024} = (1)^{2023} \

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$1:1:4$

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(D) To find $a$,the sum of all coefficients,we set $x=1$ in the expression: $a = (1-2(1)+2(1)^2)^{2023} 	imes (3-4(1)^2+2(1)^3)^{2024} = (1)^{2023} 	imes (1)^{2024} = 1$. To find $b$,we use $L'	ext{H\^opital's Rule}$ as it is a $0/0$ form: $b = \lim_{x ightarrow 0} \frac{\frac{d}{dx} \int_0^x \frac{\ln(1+t)}{t^{2024}+1} dt}{\frac{d}{dx} (x^2)} = \lim_{x ightarrow 0} \frac{\frac{\ln(1+x)}{x^{2024}+1}}{2x} = \frac{1}{2} \lim_{x ightarrow 0} \left( \frac{\ln(1+x)}{x} 	imes \frac{1}{x^{2024}+1} ight) = \frac{1}{2} 	imes 1 	imes 1 = \frac{1}{2}$. Substituting $a=1$ and $b=1/2$ into the second equation $2bx^2+ax+4=0$: $2(1/2)x^2 + (1)x + 4 = 0 \implies x^2+x+4=0$. Since the equations $cx^2+dx+e=0$ and $x^2+x+4=0$ have a common root and the coefficients are real,the ratios of the coefficients must be equal: $\frac{c}{1} = \frac{d}{1} = \frac{e}{4}$. Thus,$d:c:e = 1:1:4$.

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