Question

DAU COROANA!!!!
Exercitiul 10:a si b.
VA ROOOG!!!!

Answer 1

a) $\frac{1·2²+1}{1·2³+1·2²+1} =$$\frac{(1)(4)}{1(2³)+1(2²)+1} =$$\frac{4}{1(2³)+1(2²)+1} =$$\frac{4}{(1)(8)+1(2²)+1} =$$\frac{4}{8+1(2²)+1} =$$\frac{4}{8+(1)(4)+1} =$$\frac{4}{8+4+1} =$$\frac{4}{12+1} =$$\frac{4}{13}$

b)$\frac{1·(5²)+(3)·(5)+4}{2(4³)+3(4²)+1} = \frac{(1)(25)+(3)(5)+4}{2(4³)+3(4²)+1} = \frac{25+(3)(5)+4}{2(4³)+3(4²)+1} = \frac{25+15+4}{2(4³)+3(4²[tex] \frac{40+4}{2(4³)+3(4²)+1} = \frac{44}{2(4³)+3(4²)+1} = \frac{44}{(2)(64)+3(4²)+1} = \frac{44}{128+3(4²)+1} = \frac{44}{128+(3)(16)+1}$)+1}= [/tex]$\frac{44}{128+48+1} = \frac{44}{176+1}= \frac{44}{177}$