0.5*3^64-(3^32+1)(3^16+1)(3^8+1)(3^4+1)(3^2+1)(3+1) - вопрос №6023714053643321316138134132131 от СофияПодставкина 16.03.2023 05:10

Question

Алгебра: 0.5*3^64-(3^32+1)(3^16+1)(3^8+1)(3^4+1)(3^2+1)(3+1) решите,

СофияПодставкина · Answer

0.5*3^64-(3^32+1)(3^16+1)(3^8+1)(3^4+1)(3^2+1)(3+1)(3-1)/(3-1)=
=0.5*3^64-(3^32+1)(3^16+1)(3^8+1)(3^4+1)(3^2+1)(3^2-1)/(3-1)=
=0.5*3^64-(3^32+1)(3^16+1)(3^8+1)(3^4+1)(3^4-1)/(3-1)=
=0.5*3^64-(3^32+1)(3^16+1)(3^8+1)(3^8-1)/(3-1)=
=0.5*3^64-(3^32+1)(3^16+1)(3^16-1)/(3-1)=
=0.5*3^64-(3^32+1)(3^32-1)/(3-1)=
=0.5*3^64-(3^64-1)/(3-1)=
=0.5*3^64-(3^64-1)/2=
=0.5*3^64-0.5(3^64-1)=
=0.5*3^64-0.5*3^64+0.5=0.5