Question

(1+2+3+...+80-40X17)/10=    
3+6+...+162-27X138=    TR SA FIE PATRATE PERFECTE
   

Answer 1

[tex] \frac{1+2+3+...+80-40*17}{10} = \\ \\ \frac{ \frac{80(80+1)}{2} -40*17}{10}= \\ \\ \frac{40*81-40*17}{10}= \\ \\ \frac{40(81-17)}{10}= \\ \\ Se~simplifica~10~cu~40: \\ \\ 4*64=2^2*8^2=(8*2)^2=p.p[/tex]




[tex]3+6+...+162-27*138= \\ Dam~factor~comun~pe~3: \\ 3(1+2+3+....+54)-27*138= \\ 3* \frac{54(54+1)}{2}-27*138= \\ 3*27* 55-27*138= \\ Dam~factor~comun~pe~27: \\ 27*(3*55-138)= \\ 27*(165-138)= \\ 27*27=27^2=p.p[/tex]