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Now 81² + 25² + 14² = (9²)² + (5²)² + 14⁴ = (9² + 5²)² - 45² - 45² + 14⁴ =(9² + 5² + 45)(9² + 5² - 45) - 45² + (14²)² =(9² + 5² + 45)(9² + 5² - 45) + (14² + 45)(14² - 45) = (81 + 25 + 45)(81 + 25 - 45) + (196 + 45)(196 - 45) =(151)(61) + (241)(151) = (151)[61 + 241] = (151)(302)So, [(9²)² + 5⁴ + 14⁴]/2 =151²So, √[(81² + 25² + 14²)/2] = 151
{162+50+56}/2/=268/2=134 10^10^2^17 10^10^2^17^1 2^5^2^5^2^1^1 1^1^1^1^2 1^2 (x ➖ 2x+1).
Let y = 7, x=2(y+x)^4+(y-x)^4+16y^4 = 2(y^4 + 6x^2 y^2 + x^4) + 2(8y^4) = 2( y^4+8y^4 + 2×3×x^2 y^2 + x^4)= 2(3 y^2 +x^2)^2= 2(3×49+4)^2 = 2(151)^2
Stanford U have really dropped their standards if they use this as an entrance criterion. (They don't of course.)
Stanford this time, eh?
?? wrong from the very beginning
Now 81² + 25² + 14² =
(9²)² + (5²)² + 14⁴ =
(9² + 5²)² - 45² - 45² + 14⁴ =
(9² + 5² + 45)(9² + 5² - 45) - 45² + (14²)² =
(9² + 5² + 45)(9² + 5² - 45)
+ (14² + 45)(14² - 45) =
(81 + 25 + 45)(81 + 25 - 45)
+ (196 + 45)(196 - 45) =
(151)(61) + (241)(151) =
(151)[61 + 241] =
(151)(302)
So, [(9²)² + 5⁴ + 14⁴]/2 =
151²
So, √[(81² + 25² + 14²)/2] = 151
{162+50+56}/2/=268/2=134 10^10^2^17 10^10^2^17^1 2^5^2^5^2^1^1 1^1^1^1^2 1^2 (x ➖ 2x+1).
Let y = 7, x=2
(y+x)^4+(y-x)^4+16y^4 = 2(y^4 + 6x^2 y^2 + x^4) + 2(8y^4)
= 2( y^4+8y^4 + 2×3×x^2 y^2 + x^4)
= 2(3 y^2 +x^2)^2
= 2(3×49+4)^2 = 2(151)^2
Stanford U have really dropped their standards if they use this as an entrance criterion. (They don't of course.)
Stanford this time, eh?
?? wrong from the very beginning