ID瞩目
by DeepSkyBlue__ @ 2020-04-02 11:26:15
1
by IntrepidStrayer @ 2020-04-02 11:27:32
牛顿迭代？
by t162 @ 2020-04-02 11:27:37
不止一个答案
by minmoe @ 2020-04-02 11:28:21
@[__淦__](/user/50716) $x=1$这题不就完了吗
by tangrunxi @ 2020-04-02 11:28:42
@[__淦__](/user/50716)
$$(2x-1)^{2x}-(2x-1)=0$$
$$(2x-1)[(2x-1)^{2x-1}-1]=0$$
$$(2x-1)^{2x-1}=1$$
$$x=1$$
by Warriors_Cat @ 2020-04-02 11:28:46
1
by hanzhongtlx @ 2020-04-02 11:29:35
@[__淦__](/user/50716) 一个答案吧，我用几何画板画的
by Warriors_Cat @ 2020-04-02 11:29:52
由 $x\neq \dfrac{1}{2}$ 可得 $(2x-1)\neq0$
故可以 $(2x-1)^{2x-1}=1$
然后就不会了（
by Aw顿顿 @ 2020-04-02 11:30:36
@[__淦__](/user/50716)
by Warriors_Cat @ 2020-04-02 11:31:01