一道数学题，帮忙解答一下！谢谢！

x+1/x=3

把x^2/[x^4+x^2+1]倒过来是[x^4+x^2+1]/x^2=x^2+1/x^2+1=[x+1/x]^2-2+1=3*3-1=8

所以：x^2/[x^4+x^2+1]=1/8

[4/(x^2-y^2)+(x+y)/(x*y^2-x^2*y)]/[(x^2+xy-2*y^2)/x^2*y+2x*y^2)]
=[4/(x+y)(x-y)+(x+y)/xy(y-x)]/[(x+2y)(x-y)/xy(x+2y)]
=[(4xy-x^2-2xy-y^2)/xy(x+y)(x-y)]/[(x-y)/xy]
=[-(x-y)^2/xy(x+y)(x-y)]/[(x-y)/xy]
=-[x-y]/xy(x+y)*xy/[x-y]
=-1/[x+y]
=-1/[2-3]
=1

x^2/(x^4+x^2+1)倒过来分解为x^2+1+1/x^2=(X+1/X)^2-2+1=9-2+1=8,所以又变成倒数为原式为1/8!

√X+1/√X=根号【（x+1）/X】=根号3
（√10+X）Y=（根号10
+3）3=6+3根号10=15元

1.已知X+1/X=3，则√X+1/√X的值=根号5
2.X＝3，Y＝√10-3
所以钱包有1元

(√X+1/√X)^2=X+2+1/x
又因为X+1/X=3
所以(√X+1/√X)^2=X+2+1/x
=3+2
=5
所以√（(√X+1/√X)^2）＝√X+1/√X＝√5