a) Ta có: \(A\left(x\right)=x+x^2+...+x^{100}\)
\(\Rightarrow A\left(-1\right)=\left(-1\right)+\left(-1\right)^2+...+\left(-1\right)^{99}+\left(-1\right)^{100}\)
\(=\left(-1\right)+1+...+\left(-1\right)+1\) ( 100 số )
\(=0\)
Vậy x = -1 là nghiệm của đa thức A(x)
b) \(A\left(x\right)=x+x^2+...+x^{100}\)
\(\Rightarrow A\left(\dfrac{1}{2}\right)=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{100}\)
\(=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\)
\(\Rightarrow2A\left(\dfrac{1}{2}\right)=1+\dfrac{1}{2}+...+\dfrac{1}{2^{99}}\)
\(\Rightarrow2A\left(\dfrac{1}{2}\right)-A\left(\dfrac{1}{2}\right)=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\right)\)
\(\Rightarrow A\left(\dfrac{1}{2}\right)=1-\dfrac{1}{2^{100}}\)
Vậy khi x = \(\dfrac{1}{2}\) thì \(A=1-\dfrac{1}{2^{100}}\)
