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Автовоспроизведение
Автоповтор
E=1+2(2)^1/2
2×(2^(1/2))+1.
=1+(2)3/2
56+10√(32) = (2+2√2)^3. So, E = [25+22√2]^1/2. But 25+2√2 = (1+2√2)^3 > E = 1 +2 √2.
2√2 +1
somehow out of the "gut" the trasnformation showing there are cubes ... One can easily show by establishing a 3rd degree system of 2 vars, then noticing the (1,1) and (1,2) are the solutions for the first respective the second system ...
(2×ριζα2)+1
1+2sqrt2
{9x^3+"33x^3}= 42x^6 {168x^3+30x^3}= {198x^6+96x^3}= 294x^9 {42x^6+294x^9}=336x^15 10^306^6x^3^5 2^55^6^6^6x3^5 2^1^1^3^2^3^2^3^2x^3^1 1^1^1^1^1^1^2x^3^1 1^2x^3^1 2x^3 (x ➖ 3x+2).
cube root of 25?
E=1+2(2)^1/2
2×(2^(1/2))+1.
=1+(2)3/2
56+10√(32) = (2+2√2)^3. So, E = [25+22√2]^1/2. But 25+2√2 = (1+2√2)^3 > E = 1 +2 √2.
2√2 +1
somehow out of the "gut" the trasnformation showing there are cubes ... One can easily show by establishing a 3rd degree system of 2 vars, then noticing the (1,1) and (1,2) are the solutions for the first respective the second system ...
(2×ριζα2)+1
1+2sqrt2
{9x^3+"33x^3}= 42x^6 {168x^3+30x^3}= {198x^6+96x^3}= 294x^9 {42x^6+294x^9}=336x^15 10^306^6x^3^5 2^55^6^6^6x3^5 2^1^1^3^2^3^2^3^2x^3^1 1^1^1^1^1^1^2x^3^1 1^2x^3^1 2x^3 (x ➖ 3x+2).
cube root of 25?