已知x+1/x=3,求x^2/(x^4+x^2+1)的值!我不太会做,请大家帮帮忙,要有过程谢谢!!
还有一道:先化简后求值!
[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)]
其中x=2,y=-3,请写上过程,谢谢!!!!!!!!!!
非常感谢!!!!!!!!!
參考答案: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