Question

帮忙解个不等式 k/(1-k^2)＜1 要有详细过程.谢谢了~！

Answer 1

k/(1-k^2)＜1
k/(1-k^2) - 1 ＜ 0
{k-(1-k^2)} / (1-k^2) ＜ 0
{k^2+k-1} / (1-k^2) ＜ 0
{k^2+k-1} / (k^2-1) ＞ 0
{k^2+k-1} /正缺伏 (k^2-1) ＞ 0
{k+(1+根号5)/2} { k-(-1+根号扮历5)/2} / (k+1)(k-1)} ＞0
等效于 {k+(1+根号5)/2} {k+1} { k-(-1+根举携号5)/2} {k-1} ＞0
∴x＜-(1+根号5)/2，或-1＜x＜(-1+根号5)/2，或x＞1

Answer 2

先移项.把1移到左边同分.然后分别讨论分子分母的正负,两种情况..最后求并集就可以了.看的懂吗?

Answer 3

k/(1-k^2)<1
=k<1-k^2
=k^2+k-1<0
=k^2+k+1/枝裤4<5/4
=(k+1/2)^2<5/4
=负根号5/4<k+1/2<根仔搭伏号5/4
=负根号5/4-1/2<k<根号5/4-1/念携2