应该是1-x吧
这样则后面上下乘2^x
=1/(2^x+√2)+2^x/(2+2^x*√2)
前面上下乘√2
=√2//(2+2^x*√2)+2^x/(2+2^x*√2)
=(2^x+√2)/(2+2^x*√2)
=(2^x+√2)/[√2(2^x+√2)]
=1/√2