\(=\dfrac{2^{10}\cdot17+2^9\cdot2\cdot3\cdot5}{2^8\cdot2^3}=\dfrac{2^{10}\left(17+15\right)}{2^{11}}=\dfrac{1}{2}\cdot32=16\)
\(B=\left(4^5\cdot17+30\cdot8^3\right):\left(8\cdot16^2\right)\)
\(B=\left[\left(2^2\right)^5\cdot17+30\cdot\left(2^3\right)^3\right]:\left[2^3\cdot\left(2^4\right)^2\right]\)
\(B=\left(2^{10}\cdot17+30\cdot2^9\right):\left(2^3\cdot2^8\right)\)
\(B=\left(2^{10}\cdot17+30\cdot2^9\right):2^{11}\)
\(B=2^9\cdot\left(2\cdot17+30\cdot1\right):2^{11}\)
\(B=\dfrac{2^9}{2^{11}}\cdot\left(34+30\right)\)
\(B=\dfrac{1}{2^{11-9}}\cdot64\)
\(B=\dfrac{1}{2^2}\cdot64\)
\(B=\dfrac{64}{4}\)
\(B=16\)
Vậy B = 16
