Question

Calculați x^y+y^x pentru: x=8^33:[4^32x2^34+(2^5x2^20)^5:(16x2^23)+(7^5:7^5x1)^32x4 si y=[(11-0^11)×(3^3-3^2)+1^2020]×(3^2-2^3)-3^2×2
Va rog ,imi trebuie pana maine

Answer 1

x=8^33:[4^32x2^34+(2^5x2^20)^5:(16x2^23)]+(7^5:7^5x1)^32x4 =
=2^99:[2^64x2^34+(2^25)^5:(2^4x2^23)]+(1x1)^32x4 =
=2^99:[2^98+2^125:(2^27)]+1x4 =
=2^99:2^98(1+1)+4 =
=2^99:2^98×2+4 =
=2^99:2^99+4 =
=1+4=5

 y=[(11-0^11)×(3^3-3^2)+1^2020]×(3^2-2^3)-3^2×2 =
=[(11-0)×(3^3-3^2)+1]×(3^2-2^3)-3^2×2 =
=[11×18+1]×1-18 =
=199-18 =181

x^y+y^x=5¹⁸¹+181⁵