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49- 9= 40 gives √ x = 4 hence x= 16 +z
4:53 How do you derive a and b are integers?
40=49-9 =[sqrt(7)]⁴-[sqrt(3)]⁴Therefore sqrt(x)=4 --> x=16
Sqrt[7]^Sqrt[16]-Sqrt[3]^Sqrt[16]=40 X=16 final answer
For which class? If for the elder, then prove that the function 7^x-3^x increases monotonously - if you want algebraically, you want through a derivative. And the answer 49-9=40 is obvious.
{49 ➖ 9}=40 2^20 2^10 2^2^5 1^2^1 2^1 (x ➖ 2x+1).
Argentina, Math Olympiad: (√7)^√x - (√3)^√x = 40, x ϵ ℤ; x =?(√7)^√x - (√3)^√x = 40 = 49 - 9 = 7^2 - 3^2 = (√7)^4 - (√3)^4 = (√7)^√16 - (√3)^√16x = 16
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49- 9= 40 gives √ x = 4 hence x= 16 +z
4:53 How do you derive a and b are integers?
40=49-9
=[sqrt(7)]⁴-[sqrt(3)]⁴
Therefore sqrt(x)=4 --> x=16
Sqrt[7]^Sqrt[16]-Sqrt[3]^Sqrt[16]=40 X=16 final answer
For which class? If for the elder, then prove that the function 7^x-3^x increases monotonously - if you want algebraically, you want through a derivative. And the answer 49-9=40 is obvious.
{49 ➖ 9}=40 2^20 2^10 2^2^5 1^2^1 2^1 (x ➖ 2x+1).
Argentina, Math Olympiad: (√7)^√x - (√3)^√x = 40, x ϵ ℤ; x =?
(√7)^√x - (√3)^√x = 40 = 49 - 9 = 7^2 - 3^2 = (√7)^4 - (√3)^4 = (√7)^√16 - (√3)^√16
x = 16