Question

Answer 1

a: \(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)

\(=\frac{100+121+144}{169+196}=\frac{365}{365}=1\)

b: \(9!-8!-7!\cdot8^2\)

\(=7!\left(9\cdot8-8-8^2\right)=7!\cdot\left(72-8-64\right)\)

=0

c:\(\frac{\left(3\cdot4\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}=\frac{\left(3\cdot2^2\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot2^{22}-2^{36}}\)

\(=\frac{3^2\cdot2^{36}}{2^{35}\left(11-2\right)}=\frac{3^2}{9}\cdot2=2\)