超超簡単

作者: ママー
回答 6件
1+1=(1)﻿﻿
2+9=(2)﻿﻿
12-9=(3)﻿﻿
25×5=(4)﻿﻿
36÷9=(5)﻿﻿
8x+3=27 x=(6)﻿﻿
8x×3y-2xy=(7)﻿﻿
﻿log⁡8(64)\log_{8}\left(64\right)
log8​(64)﻿=(8)﻿﻿
﻿∫58415152 dx\int_{58}^{415}152\,dx
∫58415​152dx﻿=(9)﻿﻿
ナビエ・ストークス方程式=ρ(∂v∂t+v⋅∇v)=−∇p+μ∇2v+f\rho \left(\frac{\partial \mathbf{v}}{\partial t} + \mathbf{v} \cdot \nabla \mathbf{v}\right) = -\nabla p + \mu \nabla^2 \mathbf{v} + \mathbf{f}ρ(∂t∂v​+v⋅∇v)=−∇p+μ∇2v+f﻿﻿
ではfは何を表す? (10)﻿﻿
アインシュタインの重力場方程式=Rμν−12gμνR+Λgμν=8πGc4TμνR_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu}Rμν​−21​gμν​R+Λgμν​=c48πG​Tμν​﻿﻿
ではΛは何を表す?  (11)﻿﻿
リーマンのゼータ関数=ζ(s)=∑n=1∞1ns=∏p11−p−s\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} = \prod_{p} \frac{1}{1 - p^{-s}}
ζ(s)=∑n=1∞​ns1​=∏p​1−p−s1​﻿﻿
では
﻿∑n=1∞1ns\sum_{n=1}^{\infty} \frac{1}{n^s}∑n=1∞​ns1​﻿　は何を表す? (12)﻿﻿
﻿(x+u+v+w)(x+u−v−w)(x−u+v−w)(x−u−v+w)={(x+u)+(v+w)}{(x+u)−(v+w)}{(x−u)+(v−w)}{(x−u)−(v−w)}={(x+u)2−(v+w)2}{(x−u)2−(v−w)2}=(x+u)2(x−u)2+(v+w)2(v−w)2−(x+u)2(v−w)2−(v+w)2(x−u)2={(x+u)(x−u)}2+{(v+w)(v−w)}2−(x+u)2(v−w)2−(v+w)2(x−u)2=(x2−u2)2+(v2−w2)2−(x2+2xu+u2)(v2−2vw+w2)−(v2+2vw+w2)(x2−2xu+u2)=(x4−2u2x2+u4)+(v4−2v2w2+w4)−(v2x2−2vwx2+w2x2+2uv2x−4uvwx+2uw2x+u2v2−2u2vw+w2u2)−(v2x2−2uv2x+u2v2+2vwx2−4uvwx+2u2vw+w2x2−2uw2x+w2u2)=x4−2(u2+v2+w2)x2+8uvwx+u4+v4+w4−2(u2v2+v2w2+w2u2)=x4−2(u2+v2+w2)x2+8uvwx+(u2+v2+w2)2−4(u2v2+v2w2+w2u2)\begin{aligned}
& (x+u+v+w)(x+u-v-w)(x-u+v-w)(x-u-v+w) \\
&= \{(x+u)+(v+w)\}\{(x+u)-(v+w)\}\{(x-u)+(v-w)\}\{(x-u)-(v-w)\} \\
&= \{(x+u)^2-(v+w)^2\}\{(x-u)^2-(v-w)^2\} \\
&= (x+u)^2(x-u)^2+(v+w)^2(v-w)^2-(x+u)^2(v-w)^2-(v+w)^2(x-u)^2 \\
&= \{(x+u)(x-u)\}^2+\{(v+w)(v-w)\}^2-(x+u)^2(v-w)^2-(v+w)^2(x-u)^2 \\
&= (x^2-u^2)^2+(v^2-w^2)^2-(x^2+2xu+u^2)(v^2-2vw+w^2) \\
&\quad -(v^2+2vw+w^2)(x^2-2xu+u^2) \\
&= (x^4-2u^2x^2+u^4)+(v^4-2v^2w^2+w^4) \\
&\quad -(v^2x^2-2vwx^2+w^2x^2+2uv^2x-4uvwx+2uw^2x+u^2v^2-2u^2vw+w^2u^2) \\
&\quad -(v^2x^2-2uv^2x+u^2v^2+2vwx^2-4uvwx+2u^2vw+w^2x^2-2uw^2x+w^2u^2) \\
&= x^4-2(u^2+v^2+w^2)x^2+8uvwx+u^4+v^4+w^4-2(u^2v^2+v^2w^2+w^2u^2) \\
&= x^4-2(u^2+v^2+w^2)x^2+8uvwx+(u^2+v^2+w^2)^2-4(u^2v^2+v^2w^2+w^2u^2)
\end{aligned}
​(x+u+v+w)(x+u−v−w)(x−u+v−w)(x−u−v+w)={(x+u)+(v+w)}{(x+u)−(v+w)}{(x−u)+(v−w)}{(x−u)−(v−w)}={(x+u)2−(v+w)2}{(x−u)2−(v−w)2}=(x+u)2(x−u)2+(v+w)2(v−w)2−(x+u)2(v−w)2−(v+w)2(x−u)2={(x+u)(x−u)}2+{(v+w)(v−w)}2−(x+u)2(v−w)2−(v+w)2(x−u)2=(x2−u2)2+(v2−w2)2−(x2+2xu+u2)(v2−2vw+w2)−(v2+2vw+w2)(x2−2xu+u2)=(x4−2u2x2+u4)+(v4−2v2w2+w4)−(v2x2−2vwx2+w2x2+2uv2x−4uvwx+2uw2x+u2v2−2u2vw+w2u2)−(v2x2−2uv2x+u2v2+2vwx2−4uvwx+2u2vw+w2x2−2uw2x+w2u2)=x4−2(u2+v2+w2)x2+8uvwx+u4+v4+w4−2(u2v2+v2w2+w2u2)=x4−2(u2+v2+w2)x2+8uvwx+(u2+v2+w2)2−4(u2v2+v2w2+w2u2)​﻿なんの解法でしょう? ﻿(13)﻿﻿
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