Если в арифметической прогрессии а5=6, то найти a1+a4+a9?​

ответ: 18-разность прогрессии.

Объяснение:

1)(6ху+8у)-2ху-8у+1=6ху+8у-2ху-8у+1

2)5с+15-с^[2]-3с-3с+3с^[2]=2c+2c^[2]-6c+15

3)5b^[2]-5bc+2bc-c^[2]=5b^[2]-c^[2]-3bc

i'm not sure that's right, but folowing my minds:

as k+k=t and t+t=k it means that t=k. so it doesn't matter what number to substitute for letters, and c=0. hope, i helped you=)