Sở Tâm
1 + 1 = ?
1 + 1 = 4 – 2
Ta xét thấy: 4 = 4 + 0
=> 1 + 1 = 4 + 0 – 2
Thay đồng nhất thức Euler vào:
= 4 + (e^iπ + 1) – 2
Ta xét thấy: – 2 = 0 – 2 = 0 + (– 2)
– 2 = i² × 2
=> 1 + 1 = = 4 + (e^iπ + 1) + (i² × 2)
= (6 – 2) + (e^iπ + 1) + (i² × 2)
= (4 – 2)² + (e^iπ + 1) + (i² × 2)
Dùng hẳng đẳng thức đáng nhớ số 2: (a – b)² = a² – 2ab + b²
= (4² – 2 × 4 × 2 + 2²) + (e^iπ + 1) + (i² × 2)
= 4² – 2² × 4 + 2² + (e^iπ + 1) + (i² × 2)
= 4² – 4² + 2² + (e^iπ + 1) + (i² × 2)
Ta lại dùng hằng đẳng thức: a² – b² = (a² – b²)(a² + b²)
= (4² – 4²)(4² + 4²) + 2² + (e^iπ + 1) + (i² × 2)
= (4² – 4²)(4² + 4²) + 4 + (e^iπ + 1) + (i² × 2)
= (4² – 4²)(4² + 4²) + (6 – 2) + (e^iπ + 1) + (i² × 2)
= (4² – 4²)(4² + 4²) + (4 – 2)² + (e^iπ + 1) + (i² × 2)
= (4² – 4²)(4² + 4²) + (4² – 2 × 4 × 2 + 2²) + (e^iπ + 1) + (i² × 2)
= (4² – 4²)(4² + 4²) + 4² – 2² × 4 + 2² + (e^iπ + 1) + (i² × 2)
= (4² – 4²)(4² + 4²) + 4² – 4² + 2² + (e^iπ + 1) + (i² × 2)
= (4² – 4²)(4² + 4²) + (4² – 4²)(4² + 4²) + 2² + (e^iπ + 1) + (i² × 2)
= ((4² – 4²)(4² + 4²) × 2) + 2² + (e^iπ + 1) + (i² × 2)
= ((4² – 4²)(4² + 4²) × 2) + 4 + (e^iπ + 1) + (i² × 2)
= ((4² – 4²)(4² + 4²) × 2) + 6 – 2 + (e^iπ + 1) + (i² × 2)
= ((4² – 4²)(4² + 4²) × 2) + 6 + 0 – 2 + (e^iπ + 1) + (i² × 2)
= ((4² – 4²)(4² + 4²) × 2) + 6 + (e^iπ + 1) + (i² × 2) + (e^iπ + 1) + (i² × 2)
= ((4² – 4²)(4² + 4²) × 2) + 6 + 2(e^iπ + 1) + 2(i² × 2)
Ta dùng tính chất phân phối của phép nhân:
= ((4² – 4²)(4² + 4²) × 2) + 6 + 2((e^iπ + 1) + (i² × 2))
Ta biết: a = a × 1
=> 1 + 1 = 1((4² – 4²)(4² + 4²) × 2) + 1 × 6 + 1 × (2((e^iπ + 1) + (i² × 2)))
Ta lại dùng tính phân phối của phép cộng:
= 1(((4² – 4²)(4² + 4²) × 2) + 6 + 2((e^iπ + 1) + (i² × 2)))
= (log2(1024) – log2(16)) + 2(–2)
= ((log2(1024) – log2(16)) + 2(–2)/2) × ((–4 × i²)/2)
= ((log2(1024) – log2(16)) – (2 × 2)/2) × ((–4 × i²)/2)
= ((log2(1024) – log2(16)) – (2 × 2)/2) × ((–4 × –1)/2)
= ((log2(32) – log2(4)) – 2) × 2
= (5 – 2 – 2) × 2
= (5 – (2 + 2)) × –2 × –1
= (5 – (2 + 2)) × ((i² × 2) × (e^iπ))
= (5 – (2 + 2)) × (((i² × 2) × (cos(π) + isin(π)))
= (((( – 1, 6180339887...) × 5) – (( – 1, 6180339887...) × 2)) + ( – 1, 6180339887...) × 2)))) (((i² × 2) × (cos(π) + isin(π)))
= (0 – ((((( – 1, 6180339887...) × 5) – (( – 1, 6180339887...) × 2)) + ( – 1, 6180339887...) × 2)))) (((i² × 2) × (cos(π) + isin(π))))) × i²
= ((cos(π) + isin(π)) – ((((( – 1, 6180339887...) × 5) – (( – 1, 6180339887...) × 2)) + ( – 1, 6180339887...) × 2)))) (((i² × 2) × (cos(π) + isin(π))))) × i²
= F(x)
(Với x = ((cos(π) + isin(π)) – ((((( – 1, 6180339887...) × 5) – (( – 1, 6180339887...) × 2)) + ( – 1, 6180339887...) × 2)))) (((i² × 2) × (cos(π) + isin(π))))) × i²)
= F(y) × F(z)
(Với y = ((cos(π) + isin(π)) – (((((\phi – 0, 6180339887...) × 5) – ((\phi – 0, 6180339887...) × 2)) + (\phi – 0, 6180339887...) × 2)))) (((i² × 2) × (cos(π) + isin(π))))) và z = i²)
(Thay đổi, với y = ((cos(π) + isin(π)) – ((((((1 + √5/2) – 0, 6180339887...) × 5) – (((1 + √5/2) – 0, 6180339887...) × 2)) + ((1 + √5/2) – 0, 6180339887...) × 3)))) (((i² × 2) × (cos(π) + isin(π)))))
= F(y) × F(z)
Tính toán số 2.
= √2²
= d/dx(2x)
= \int–0^2,dx
= ||(1, 1)||²
= det ( 1 0 0 2)
= limx→0sin(2x)/sin(x)
= ln(e²)
= 2!
= 2³ – 2² – 2¹
= 2 × 2 × 2 – 2 × 2 – 2
= (2 × 2 × 2) – (2 × 2) – (2)
= ln(e²)(((√2²) × (d/dx(2x)) × (\int–0^2,dx)) – ((||(1, 1)||²) × (det ( 1 0 0 2))) – (limx→0sin(2x)/sin(x)))/2!
=> Cuối cùng là...
= e^iπ(ln(e²)(((√2²) × (d/dx(2x)) × (\int–0^2,dx)) – ((||(1, 1)||²) × (det ( 1 0 0 2))) – (limx→0sin(2x)/sin(x)))/2!)/i² × ((\phi – 0, 6180339887...) × log2(2) × F(n))
(Với F(n), n = 1)
= F(y) × F(z) × F(n)
= (∑(n=1→∞)(1/2)^n) × ((e^0 + (P(A)+P(A^c))) / tr(1 0 0 1))
× |1+i|² × ((√2)²/C_2^1) × (Γ(3)/((1/2)×1×2²))
= 1 × ((1+1)/2) × 2 × (2/2) × (2/2)
= 1 × 1 × 2 × 1 × 1
= 2.
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