a. A= \(100^2-99^2+98^2-97^2+...+2^2-1^2\)

\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)

\(=1\left(100+99\right)+1\left(98+97\right)+...+1\left(2+1\right)\)

\(=100+99+98+97+...+2+1 \\ =\left(100+1\right).100:2\\ =5050\)

b.B=\(3.\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)

\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)

\(=\left(2^8-1\right)...\left(2^{64}+1\right)+1^2\)

\(=\left(2^{64}-1\right)\left(2^{64}+1\right)+1^2\)

\(=2^{128}-1+1 \\ =2^{128}\)

Ta có:
\(\dfrac{x}{x-4}=\dfrac{...}{x^2-16}=\dfrac{...}{\left(x-4\right)\left(x+4\right)}\)

\(\Rightarrow\dfrac{...}{\left(x-4\right)\left(x+4\right)}=\dfrac{x}{x-4}=\dfrac{x\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{x^2+4x}{\left(x-4\right)\left(x+4\right)}\)

Vậy đa thức cần điền vào dấu ... là \(x^2+4x\)