Question

Rezolvati in N ecuatia:
$A^{y} _{x} = x+y$

Answer 1

Răspuns:

Explicație pas cu pas:

Aₓⁿ = x + n

Conditii :   x, n ∈ N ; x ≥ n

Aₓⁿ = x + n   <=> x! / (x-n)! = x + n

<=> x! = (x+n)·(x-n)!  <=>

x·(x-1)·(x-2)·....(x-n)! = (x+n)·(x-n)! =>

x·(x-1)·......(x-n+1) = x+n

n = 0 => Aₓ⁰ = x+0  <=> x! / x! = x  <=> x = 1

n = 2 => x(x-1) = x+2  => x²-x = x + 2  => x²-2x-2 = 0 =>

x₁,₂ = [2±√(4+8)]/2 ∉N

n = 3 => x(x-1)(x-2) = x+3 =>(x²-x)(x-2) = x+3 =>

x³-3x²+x-3 = 0  <=> x²(x-3)+(x-3) = 0 <=>(x-3)(x²+1) = 0 => x = 3

n = 4 => x(x-1)(x-2)(x-3) = x+4 <=>

(x²-x)(x²-5x+3) = x+4 => x⁴-6x³+8x²-4x-4 = 0 => x ∉ N (x <0)

n = 5  => x(x-1)(x-2)(x-3)(x-4) = x+5  <=>

(x²-x)(x²-5x+6)(x-4)-(x+5) = 0  <=>

(x³-5x²+4x)(x²-5x+6)-(x+5) = 0 =>

x⁵-10x⁴+35x³-50x²+23x-5 = 0 => x < 0 ; x ∉ N

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Solutii :

x = 1  ; n = 0

x = n = 3

Answer 2

$A_x^y = x+y\Rightarrow \dfrac{x!}{(x-y)!} = x+y \Rightarrow\\ \\ \Rightarrow \dfrac{(x-y)!(x-y+1)(x-y+2)\cdot ...\cdot (x-y+y)}{(x-y)!} = x+y\\ \\ \bullet \, y \geq 1:\\ \\\Rightarrow x(x-1)(x-2)\cdot ...\cdot \big(x-(y-1)\big) = x+y\Big|:(x\neq 0)\\ \\ \bullet \, y \geq 2:\\ \\ \Rightarrow (x-1)(x-2)\cdot ...\cdot \big(x-(y-1)\big) = 1+\dfrac{y}{x}\\ \\ \text{Deoarece }x \,| \,y \,\text{ si }\,x\geq y \Rightarrow x=y\\ \\ \Rightarrow (x-1)! = 1+\frac{x}{x}\Rightarrow (x-1)! = 2 \Rightarrow x=3\\ \\ \Rightarrow (x,y) = (3,3)$

$\\\textbf{Cazul }y=1:$

$\Rightarrow A_{x}^1 = x+1 \Rightarrow \dfrac{x!}{(x-1)!} = x+1\Rightarrow\\ \\ \Rightarrow \dfrac{(x-1)!\cdot x}{(x-1)!} = x+1 \Rightarrow x = x+1 \Rightarrow 0=1\,\,(F)$

$\\\textbf{Cazul }y = 0:$

$A_{x}^0 = x+0 \Rightarrow 1 = x \Rightarrow x = 1\\ \\ \Rightarrow (x,y) =(1,0)$

$\\\text{Din cazurile }(y=0), (y=1)\text{ sau } (y\geq 2):$

$\Rightarrow \boxed{(x,y) \in\left\{(1,0); (3,3)\right\}}$