Forgot password?
 Create new account
Search
View: 94|Reply: 4

[函数] 根式化简

[Copy link]

68

Threads

434

Posts

4269

Credits

Credits
4269

Show all posts

hejoseph Post time 2023-1-16 14:20 |Read mode
本帖最后由 hejoseph 于 2023-1-16 15:01 编辑 已知 $0\leqslant a\leqslant b$,$0\leqslant a\leqslant c$,$2(ab+ac+bc)\geqslant a^2+b^2+c^2$,化简
\begin{align*}
&\sqrt{-2b^2+b(c+a)-(c-a)^2+2b\sqrt{a^2+b^2+c^2-ab-ac-bc}}\\
&{}+\sqrt{-2c^2+c(a+b)-(a-b)^2+2c\sqrt{a^2+b^2+c^2-ab-ac-bc}}
\end{align*}

730

Threads

110K

Posts

910K

Credits

Credits
93648
QQ

Show all posts

kuing Post time 2023-1-16 15:44
当 `a` 不是最小的时候似乎也能化简,结果有所不同,是
\[\sqrt{-2a^2-5b^2-5c^2+5a(b+c)+2bc+(4b+4c-2a)\sqrt{a^2+b^2+c^2-ab-ac-bc}}.\]

68

Threads

434

Posts

4269

Credits

Credits
4269

Show all posts

 Author| hejoseph Post time 2023-1-16 15:59
kuing 发表于 2023-1-16 15:44
当 `a` 不是最小的时候似乎也能化简,结果有所不同,是
\[\sqrt{-2a^2-5b^2-5c^2+5a(b+c)+2bc+(4b+4c-2a)\s ...

是的,只是结论不那么好看

68

Threads

434

Posts

4269

Credits

Credits
4269

Show all posts

 Author| hejoseph Post time 2023-2-20 14:12
无人做,直接发结论,化简结果是
\[
\sqrt{-2a^2+a(b+c)-(b-c)^2+2a\sqrt{a^2+b^2+c^2-ab-ac-bc}}
\]

3151

Threads

8383

Posts

610K

Credits

Credits
65397
QQ

Show all posts

hbghlyj Post time 2023-9-15 18:17
如何化简呢

手机版|悠闲数学娱乐论坛(第3版)

2025-3-6 21:50 GMT+8

Powered by Discuz!

× Quick Reply To Top Return to the list