Question

y=(10^x)-(10^-x)/(10^x)+(10^-x) + 1

Answer 1

楼主你好令10^x=t,则y={[t-(1/t)]/[t+(1/t)]}+1=2t^2/(t^2+1),所以t^2y+y=2t^2,所以(2-y)t^2=y,知道y≠2,所以t^2=y/(2-y),即100^x=y/(2-y),所以x=log100(100^x)=log100(y/(2-y)),所以反函数是y=log100(x/(2-x))希望...