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Math Olympiad Question: a + 3125ᵃ = 0; a = ?First method:a = - 3125ᵃ, a² = (- 3125ᵃ)² = (5⁵ᵃ)² = 5¹⁰ᵃ, (a²)¹⸍²ᵃ = (5¹⁰ᵃ)¹⸍²ᵃ, a¹⸍ᵃ = 5⁵a¹⸍ᵃ = (a⁻¹)⁻⁽¹⸍ᵃ⁾ = (- 1/a)⁽⁻¹⸍ᵃ⁾ = 5⁵, - 1/a = 5; a = - 1/5Second method:- a = 3125ᵃ, (- a)¹⸍ᵃ = (3125ᵃ)¹⸍ᵃ, (- a)¹⸍ᵃ = (- a)⁻ᵃ = 3125 = 5⁵(- a)⁻ᵃ = 5⁵ = 5⁽⁻¹⁾⁽⁻⁵⁾ = (5⁻¹)⁻⁵ = (1/5)¹⸍⁽¹⸍⁵⁾; - a = 1/5; a = - 1/5Answer check:a + 3125ᵃ = - 1/5 + (5⁵)⁽⁻¹⸍⁵⁾ = - 1/5 + 5⁻¹ = - 1/5 + 1/5 = 0Final answer:a = - 1/5
This question make sense only when a1/e, and for y>1,it has only one solution for x, so b=1/5 is the only solution.
More Faster: a
very nice.
the number 3125 gives the methodology obvious.its 5 with exponent 5!!Thats the clue and a teaching moment.
VER EL VIDEO ENTERO
Yes. They should watch over and over again to master the two most powerful 💪 methods presented.
Math Olympiad Question: a + 3125ᵃ = 0; a = ?
First method:
a = - 3125ᵃ, a² = (- 3125ᵃ)² = (5⁵ᵃ)² = 5¹⁰ᵃ, (a²)¹⸍²ᵃ = (5¹⁰ᵃ)¹⸍²ᵃ, a¹⸍ᵃ = 5⁵
a¹⸍ᵃ = (a⁻¹)⁻⁽¹⸍ᵃ⁾ = (- 1/a)⁽⁻¹⸍ᵃ⁾ = 5⁵, - 1/a = 5; a = - 1/5
Second method:
- a = 3125ᵃ, (- a)¹⸍ᵃ = (3125ᵃ)¹⸍ᵃ, (- a)¹⸍ᵃ = (- a)⁻ᵃ = 3125 = 5⁵
(- a)⁻ᵃ = 5⁵ = 5⁽⁻¹⁾⁽⁻⁵⁾ = (5⁻¹)⁻⁵ = (1/5)¹⸍⁽¹⸍⁵⁾; - a = 1/5; a = - 1/5
Answer check:
a + 3125ᵃ = - 1/5 + (5⁵)⁽⁻¹⸍⁵⁾ = - 1/5 + 5⁻¹ = - 1/5 + 1/5 = 0
Final answer:
a = - 1/5
This question make sense only when a1/e, and for y>1,it has only one solution for x, so b=1/5 is the only solution.
More Faster: a
very nice.
the number 3125 gives the methodology obvious.its 5 with exponent 5!!Thats the clue and a teaching moment.
VER EL VIDEO ENTERO
Yes. They should watch over and over again to master the two most powerful 💪 methods presented.