宇宙级数学难题

我来解答一下吧:把已知条件变为:x^(0.5)+x^(-0.5)=3
首先解分母:x^2+x^(-2)+3=[x+x^(-1)]^2-2+3=[x+x^(-1)]^2+1
而x+x^(-1)=[x^(0.5)+x^(-0.5)]^(2)-2=7
所以由上式可得分母的值:7^(2)+1=50
下面再来做分子:x^(3/2)+x^(-3/2)=[x^(0.5)]^3+[x^(-0.5)]^3=[x^(0.5)+x^(-0.5)][x+x^(-1)-1]=3*6=18
所以分子为20，所以最后原式的值:0.4即5分之2

x^(0.5)+1/x^(0.5)=3则x+1/x=3×3-2=7，进而x^2+x^(-2)+3=7×7-2+3=50
（x+1/x）×[x^(0.5)+1/x^(0.5)]=x^(3/2)+x^(-3/2)+x^(0.5)+1/x^(0.5)=x^(3/2)+x^(-3/2)+3=7×3=21，所以x^(3/2)+x^(-3/2)+2=21-3+2=20
原式=20/50=0.4。

解:
(x^(3/2)+x^(-3/2)+2)/(x^2+x^(-2)+3)
=(x^(3/2)+1/x^(3/2)+2)/(x^2+1/x^2+3)
利用立方和公式
=[(x^1/2+1/x^1/2)*(x+1/x -1)+2]/[(x+1/x)^2+1]
因为:x^(0.5)+1/x^(0.5)=3
所以:x+1/x=3^2-2=7
=3*(x+1/x -1)+2]/[(x+1/x)^2+1]
=20/50
=2/5
=12/25

x^(0.5)+1/x^(0.5)=3
[x^(0.5)+1/x^(0.5)]^2=3^2
x+2+x^(-1)=9
x+x^(-1)=7
[x^(3/2)+x^(-3/2)+2]/[x^2+x^(-2)+3]
={[x^1/2+x^(-1/2)]*[x-1+x^(-1)]+2}/{[x+x^(-1)]^2+1}
=[3*(7-1)+2]/50
=20/50