1/ Thay $x=1$ vào $B(x)=-4x^2+x-\dfrac{1}{4}$
$B(x)=-4.1^2+1-\dfrac{1}{4}\\\quad\quad\,\,\,=-4+1-\dfrac{1}{4}\\\quad\quad\,\,\,=-\dfrac{13}{4}$
Vậy $B(x)=-\dfrac{13}{4}$ tại $x=1$
2/ $A(x)=-x^3+2x^3+5x^2-5x^5-x^3+\dfrac{1}{4}+5x^5-2x\\\quad\quad\,\,\, =(-x^3+2x^3-x^3)+(-5x^5-5x^5)+5x^2-2x+\dfrac{1}{4}\\\quad\quad\,\,\, =5x^2-2x+\dfrac{1}{4}$
Vậy $A(x)=5x^2-2x+\dfrac{1}{4}$
3/ $P(x)=A(x)+B(x)\\\quad\quad\,\,\,=\left(5x^2-2x+\dfrac{1}{4}\right)+\left(-4x^2+x-\dfrac{1}{4}\right)\\\quad\quad\,\,\, =5x^2-2x+\dfrac{1}{4}-4x^2+x-\dfrac{1}{4}\\\quad\quad\,\,\,=(5x^2-4x^2)+(-2x+x)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)\\\quad\quad\,\,\,=x^2-x$
$P(x)=0\\↔x^2-x=0\\↔x(x-1)=0\\↔\left[\begin{array}{1}x=0\\x-1=0\end{array}\right.\\↔\left[\begin{array}{1}x=0\\x=1\end{array}\right.$
Vậy $x∈\{0;1\}$ thì $P(x)=0$
