a/ ĐKXĐ : \(\left\{{}\begin{matrix}z\ne0\\z\ne4,-4\end{matrix}\right.\)
Thay \(z=1\) vào biểu thức A ta có :
\(A=\frac{2.1}{1^2-4}=\frac{2}{-3}=-\frac{2}{3}\)
Vậy....
b/ Ta có :
\(B=\frac{2z^3+4z}{z^4-8z^2+16}\)
\(=\frac{2z\left(z+2\right)}{\left(z^2-4\right)^2}\)
\(=\frac{2z\left(z+2\right)}{\left(z-2\right)^2\left(z+2\right)^2}\)
\(=\frac{2z}{\left(z-2\right)^2\left(z+2\right)}\)
Lại có : \(M=A:B\)
\(\Leftrightarrow M=\frac{2z}{\left(z-2\right)\left(z+2\right)}:\frac{2z}{\left(z-2\right)^2\left(z+2\right)}\)
\(=\frac{2z}{\left(z-2\right)\left(z+2\right)}.\frac{\left(z-2\right)^2\left(z+2\right)}{2z}\)
\(=z-2\)
Vậy...