Выполните действия 1)5/b+2 - 1/b 2) x^2-3xy/x^2-y^2 + y/x-y только подробно ! - id601530720210515 от Mozgovoi1 15.05.2021 19:06

Question

Алгебра: Выполните действия 1)5/b+2 - 1/b 2) x^2-3xy/x^2-y^2 + y/x-y только подробно !

Mozgovoi1 · Accepted Answer

1/5*6^1024-[(6^512+1)(6^256+1)(6^128+1)(6^64+1)(6^32+1)(6^16+1)(6^8+1)(6^4+1)(6²+1)(6+1)(6-1)]/(6-1)=1/5*6^1024-1/5[(6^512+1)(6^256+1)(6^128+1)(6^64+1)(6^32+1)(6^16+1)(6^8+1)(6^4+1)(6²+1)(6²-1)]=
=1/5*6^1024-1/5[(6^512+1)(6^256+1)(6^128+1)(6^64+1)(6^32+1)(6^16+1)(6^8+1)(6^4+1)(6^4-1)]=1/5*6^1024-1/5[(6^512+1)(6^256+1)(6^128+1)(6^64+1)(6^32+1)(6^16+1)(6^8+1)(6^8-1)]=1/5*6^1024-1/5[(6^512+1)(6^256+1)(6^128+1)(6^64+1)(6^32+1)(6^16+1)(6^16-1)=1/5*6^1024-1/5[(6^512+1)(6^256+1)(6^128+1)(6^64+1)(6^32+1)(6^32-1)]=1/5*6^1024-1/5[(6^512+1)(6^256+1)(6^128+1)(6^64+1)(6^64-1)]=1/5*6^1024-1/5[(6^512+1)(6^256+1)(6^128+1)(6^128-1)]=1/5*6^1024-1/5[(6^512+1)(6^256+1)(6^256-1)]=1/5*6^1024-1/5[(6^512+1)(6^512-1)]=1/5*6^1024-1/5(6^1024-1)=1/5*6^1024-1/5*6^1024+1/5=0,2

Выполните действия 1)5/b+2 - 1/b 2) x^2-3xy/x^2-y^2 + y/x-y только подробно !

MOGZ ответил