Математика! (1/2)^4n+1=1/√2

1) http://bit.ly/2jYIeOX2) http://bit.ly/2kWg7xh3) http://bit.ly/2krehIr4) http://bit.ly/2kr9WERОтветы будут равны:n1 = log(-2^(3/4)/(2*(-sqrt(2) + 2)^(1/4)) + 2^(3/4)*i/(2*(-sqrt(2) + 2)^(1/4)))/log(2)n2 = log(2^(3/4)/(2*(-sqrt(2) + 2)^(1/4)) - 2^(3/4)*i/(2*(-sqrt(2) + 2)^(1/4)))/log(2)n3 = log(2^(3/4)/(2*(-sqrt(2) + 2)^(1/4)) + 2^(3/4)*i/(2*(-sqrt(2) + 2)^(1/4)))/log(2)n4 = log(-2^(3/4)/(2*(-sqrt(2) + 2)^(1/4)) - 2^(3/4)*i/(2*(-sqrt(2) + 2)^(1/4)))/log(2)