已知x/(x^2-x+1)=7,求x^2/(x^4+x^2+1)的值.):∵x/(x^2-x+1)=7,∴x/x(x-1+1/x)=7,∴1/(x-1+1/x)=7,∴x-1+1/x=1/7,∴x+1/x=8/7,∴(x+1/x)^2=x^2+2+1/x^2=64/49∴x^2+1+1/x^2=15/49,∴x^2/(x^4+x^2+1)=x^2/x^2(x^2+1+1/x^2)=1/(x^2+1+1/x^2)=49/15,错处
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已知x/(x^2-x+1)=7,求x^2/(x^4+x^2+1)的值.):∵x/(x^2-x+1)=7,∴x/x(x-1+1/x)=7,∴1/(x-1+1/x)=7,∴x-1+1/x=1/7,∴x+1/x=8/7,∴(x+1/x)^2=x^2+2+1/x^2=64/49∴x^2+1+1/x^2=15/49,∴x^2/(x^4+x^2+1)=x^2/x^2(x^2+1+1/x^2)=1/(x^2+1+1/x^2)=49/15,错处
已知x/(x^2-x+1)=7,求x^2/(x^4+x^2+1)的值.):∵x/(x^2-x+1)=7,∴x/x(x-1+1/x)=7,∴1/(x-1+1/x)=7,∴x-1+1/x=1/7,∴x+1/x=8/7,∴(x+1/x)^2=x^2+2+1/x^2=64/49∴x^2+1+1/x^2=15/49,∴x^2/(x^4+x^2+1)=x^2/x^2(x^2+1+1/x^2)=1/(x^2+1+1/x^2)=49/15,错处:∴(x+1/x)^2=x^2+2+1/x^2=64/49∴x^2+1+1/x^2=15/49.未去分母的分式不能直接在分子上减1!我个人认为是错误的解法,请问这样解是不是错了!
你们都做错了
已知x/(x^2-x+1)=7,求x^2/(x^4+x^2+1)的值.):∵x/(x^2-x+1)=7,∴x/x(x-1+1/x)=7,∴1/(x-1+1/x)=7,∴x-1+1/x=1/7,∴x+1/x=8/7,∴(x+1/x)^2=x^2+2+1/x^2=64/49∴x^2+1+1/x^2=15/49,∴x^2/(x^4+x^2+1)=x^2/x^2(x^2+1+1/x^2)=1/(x^2+1+1/x^2)=49/15,错处
没错啊,兄弟,只不过你导了半天把结果又倒回去了!
其中
∵x/(x^2-x+1)=7,∴x/x(x-1+1/x)=7,∴1/(x-1+1/x)=7,∴x-1+1/x=1/7,∴x+1/x=8/7,∴(x+1/x)^2=x^2+2+1/x^2=64/49 {∴x^2+1+1/x^2=15/49,} ∴x^2/(x^4+x^2+1)=x^2/x^2(x^2+1+1/x^2)=1/(x^2+1+1/x^2)=49/15, .【∴x^2+1+1/x^2=15/49,】这不就又回去了么?