Question

calculati a=(b-c)^2011,unde b=[(2^3)^2-1954^0]:3^2^1^7-(4^1^2^3-1^4^3^2)
iar c=32x7^5-14*5+3

Answer 1

Atentie!

In enuntul problemei s-a strecurat o greseala!

Enuntul corect este:

Calculati a=(b-c)^2011,

unde b=[(2^3)^2-1954^0]:3^2^1^7-(4^1^2^3-1^4^3^2)

iar c=32x7^5-14^5+3

Rezolvare:

a=(b-c)^2011=

={[(2^3)^2-1954^0]:3^2^1^7-(4^1^2^3-1^4^3^2)-(32x7^5-14^5+3)}^2011=

pentru ca: 1954^0=1; 3^2^1^7=9; 4^1^2^3=4; 1^4^3^2=1

=(2^6-1):3^2-(4-1)-(32*7^5-14^5+3)^2011=

=(2^6-1):3^2-(4-1)-(32*7^5-7*2^5+3)^2011=

=63:9-3-(2^5*7^5-(7*2)^5+3)^2011=

=7-3-(2^5*7^5-2^5*7^5+3)^2011=

=(4-3)^2011=1^2011=1