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(2)+(2)=4 (y ➖ 2x+2) (232)+(232)(1^1^1)+(1^3^2) (y ➖ 3x+2).
(x + y)² = x² + y² + 2xy → given: x + y = 416 = x² + y² + 2xy(x + y)³ = (x + y)².(x + y)(x + y)³ = (x² + 2xy + y²).(x + y)(x + y)³ = x³ + x²y + 2x²y + 2xy² + xy² + y³(x + y)³ = x³ + y³ + 3x²y + 3xy²x³ + y³ = (x + y)³ - 3x²y - 3xy²x³ + y³ = (x + y)³ - 3xy.(x + y) → given: x + y = 4x³ + y³ = 64 - 12xy ← equation (1)(x + y)⁵ = (x + y)².(x + y)².(x + y)(x + y)⁵ = (x² + 2xy + y²).(x² + 2xy + y²).(x + y)(x + y)⁵ = (x⁴ + 2x³y + x²y² + 2x³y + 4x²y² + 2xy³ + x²y² + 2xy³ + y⁴).(x + y) (x + y)⁵ = (x⁴ + 6x²y² + 4x³y + 4xy³ + y⁴).(x + y) (x + y)⁵ = x⁵ + x⁴y + 6x³y² + 6x²y³ + 4x⁴y + 4x³y² + 4x²y³ + 4xy⁴ + xy⁴ + y⁵ (x + y)⁵ = x⁵ + y⁵ + 5x⁴y + 5xy⁴ + 10x³y² + 10x²y³x⁵ + y⁵ = (x + y)⁵ - 5x⁴y - 5xy⁴ - 10x³y² - 10x²y³x⁵ + y⁵ = (x + y)⁵ - 5xy.(x³ + y³) - 10x²y².(x + y) → given: x + y = 4x⁵ + y⁵ = 1024 - 5xy.(x³ + y³) - 40x²y² → recall (1): x³ + y³ = 64 - 12xyx⁵ + y⁵ = 1024 - 5xy.(64 - 12xy) - 40x²y²x⁵ + y⁵ = 1024 - 320xy + 60x²y² - 40x²y²x⁵ + y⁵ = 1024 - 320xy + 20x²y² → given: x⁵ + y⁵ = 464464 = 1024 - 320xy + 20x²y²20x²y² - 320xy + 560 = 0x²y² - 16xy + 28 = 0 → let: z = xyz² - 16z + 28 = 0Δ = (- 16)² - (4 * 28) = 256 - 112 = 144 = 12²z = (16 ± 12)/2z = 8 ± 6 → recall: z = xyxy = 8 ± 6 ← this is the product Px + y = 4 ← this is the sum SSo x & y are the solution of the following equation: a² - Sa + P = 0First case: xy = 14 and (x + y) = 4a² - 4a + 14 = 0Δ = (- 4)² - (4 * 14) = 16 - 56 = - 40 = 40i²a = (4 ± i√40)/2 a = (4 ± 2i√10)/2 a = 2 ± i√10First solution: x = 2 + i√10 → y = 2 - i√10Second solution: x = 2 - i√10 → y = 2 + i√10Second case: xy = 2 and (x + y) = 4a² - 4a + 2 = 0Δ = (- 4)² - (4 * 2) = 16 - 8 = 8a = (4 ± √8)/2 a = (4 ± 2√2)/2 a = 2 ± √2Third solution: x = 2 + √2 → y = 2 - √2Fourth solution: x = 2 - √2 → y = 2 + √2
Excellent delivery! 👏
you never explained why you selected (x+y)^4 for expansion.
I used it because it has an expansion similar to x^5+y^5 property.
(2)+(2)=4 (y ➖ 2x+2) (232)+(232)(1^1^1)+(1^3^2) (y ➖ 3x+2).
(x + y)² = x² + y² + 2xy → given: x + y = 4
16 = x² + y² + 2xy
(x + y)³ = (x + y)².(x + y)
(x + y)³ = (x² + 2xy + y²).(x + y)
(x + y)³ = x³ + x²y + 2x²y + 2xy² + xy² + y³
(x + y)³ = x³ + y³ + 3x²y + 3xy²
x³ + y³ = (x + y)³ - 3x²y - 3xy²
x³ + y³ = (x + y)³ - 3xy.(x + y) → given: x + y = 4
x³ + y³ = 64 - 12xy ← equation (1)
(x + y)⁵ = (x + y)².(x + y)².(x + y)
(x + y)⁵ = (x² + 2xy + y²).(x² + 2xy + y²).(x + y)
(x + y)⁵ = (x⁴ + 2x³y + x²y² + 2x³y + 4x²y² + 2xy³ + x²y² + 2xy³ + y⁴).(x + y)
(x + y)⁵ = (x⁴ + 6x²y² + 4x³y + 4xy³ + y⁴).(x + y)
(x + y)⁵ = x⁵ + x⁴y + 6x³y² + 6x²y³ + 4x⁴y + 4x³y² + 4x²y³ + 4xy⁴ + xy⁴ + y⁵
(x + y)⁵ = x⁵ + y⁵ + 5x⁴y + 5xy⁴ + 10x³y² + 10x²y³
x⁵ + y⁵ = (x + y)⁵ - 5x⁴y - 5xy⁴ - 10x³y² - 10x²y³
x⁵ + y⁵ = (x + y)⁵ - 5xy.(x³ + y³) - 10x²y².(x + y) → given: x + y = 4
x⁵ + y⁵ = 1024 - 5xy.(x³ + y³) - 40x²y² → recall (1): x³ + y³ = 64 - 12xy
x⁵ + y⁵ = 1024 - 5xy.(64 - 12xy) - 40x²y²
x⁵ + y⁵ = 1024 - 320xy + 60x²y² - 40x²y²
x⁵ + y⁵ = 1024 - 320xy + 20x²y² → given: x⁵ + y⁵ = 464
464 = 1024 - 320xy + 20x²y²
20x²y² - 320xy + 560 = 0
x²y² - 16xy + 28 = 0 → let: z = xy
z² - 16z + 28 = 0
Δ = (- 16)² - (4 * 28) = 256 - 112 = 144 = 12²
z = (16 ± 12)/2
z = 8 ± 6 → recall: z = xy
xy = 8 ± 6 ← this is the product P
x + y = 4 ← this is the sum S
So x & y are the solution of the following equation: a² - Sa + P = 0
First case: xy = 14 and (x + y) = 4
a² - 4a + 14 = 0
Δ = (- 4)² - (4 * 14) = 16 - 56 = - 40 = 40i²
a = (4 ± i√40)/2
a = (4 ± 2i√10)/2
a = 2 ± i√10
First solution: x = 2 + i√10 → y = 2 - i√10
Second solution: x = 2 - i√10 → y = 2 + i√10
Second case: xy = 2 and (x + y) = 4
a² - 4a + 2 = 0
Δ = (- 4)² - (4 * 2) = 16 - 8 = 8
a = (4 ± √8)/2
a = (4 ± 2√2)/2
a = 2 ± √2
Third solution: x = 2 + √2 → y = 2 - √2
Fourth solution: x = 2 - √2 → y = 2 + √2
Excellent delivery! 👏
you never explained why you selected (x+y)^4 for expansion.
I used it because it has an expansion similar to x^5+y^5 property.