Question

(10²+11²+12³)÷(13²+14²)

Answer 1

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

\(=\left(100+121+144\right):\left(169+196\right)\)

\(=365:365\)

\(=1\)

Answer 2

\(\left(10^2+11^2+12^3\right):\left(13^2+14^2\right)\)

\(=\left(100+121+1728\right):\left(169+196\right)\)

\(=1949:365\)

\(=\dfrac{1949}{365}\)