A nice mathematics algebra exponential problem | Olympiad Question| x=?
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- เผยแพร่เมื่อ 20 ก.ย. 2024
- #quantativeaptitude #maths #olympiad #exponents #integral#rational
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_a = x/(x + 1), b = x/(x - 1)_
⇒ _1/a + 1/b = 2x/x = 2_
Multiply through by _ab_ :
_a + b = 2ab_
∴ _(a + b)² = a² + b² + 2ab = 12 + (a + b)_
⇒ _(a + b) = 4_ or _-3_
⇒ _2x²/(x² - 1) = 4_ or _-3_
⇒ *_x = ±√2_** or **_±√(3/5)_*
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(x^2/{x^2+2})=+(x^2/x^2 ➖ 1) =x^2/2x^2+x^2/{x^0+x^0 ➖ }={x^2/2x^2+x^2/x^1}=x^4/2x^3 =2,x^1.1 2x^1(x ➖ 2x+1).
[x/(x + 1)]² + [x/(x - 1)]² = 12 → where: x ≠ ± 1
[x²/(x + 1)²] + [x²/(x - 1)²] = 12
[x².(x - 1)² + x².(x + 1)²]/[(x + 1)².(x - 1)²] = 12
x².(x - 1)² + x².(x + 1)² = 12.(x + 1)².(x - 1)²
x².[(x - 1)² + (x + 1)²] = 12.[(x + 1).(x - 1)]²
x².[x² - 2x + 1 + x² + 2x + 1] = 12.(x² - 1)²
x².(2x² + 2) = 12.(x² - 1)²
2x².(x² + 1) = 12.(x² - 1)²
x².(x² + 1) = 6.(x² - 1)²
x⁴ + x² = 6.(x⁴ - 2x² + 1)
x⁴ + x² = 6x⁴ - 12x² + 6
5x⁴ - 13x² + 6 = 0
x⁴ - (13/5).x² + (6/5) = 0 → is it a square ? perhaps… (a - b)² = a - 2ab + b²
x⁴ - [2 * (13/10) * x²] + (6/5) = 0
x⁴ - [2 * x² * (13/10)] + (13/10)² - (13/10)² + (6/5) = 0
x⁴ - [2 * x² * (13/10)] + (13/10)² - (169/100) + (120/100) = 0
x⁴ - [2 * x² * (13/10)] + (13/10)² - (49/100) = 0
x⁴ - [2 * x² * (13/10)] + (13/10)² = 49/100
[x² - (13/10)]² = (7/10)²
x² - (13/10) = ± (7/10)
x² = (13/10) ± (7/10)
x² = (13 ± 7)/10
First case: x² = (13 + 7)/10 = 2
x² = 2
x = ± √2
Second case: x² = (13 - 7)/10 = 6/10 = 3/5
x² = 3/5
x = ± √(3/5)
x = ± (√3)/(√5)
x = ± (√3).(√5)/5
x = ± (√15)/5