[tex]\it\ Din\ A_{x-2}^5 \Rightarrow x-2\geq5 \Rightarrow x\geq7 ,\ x\in \mathbb{N} \ \ \ \ (1)
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A_{x-2}^5 =\dfrac{(x-2)!}{(x-2-5)!} =\dfrac{(x-2)!}{(x-7)!}[/tex]
Inecuația devine:
[tex]\it 6(x-5)!\dot\ \dfrac{(x-2)!}{(x-7)!} \leq x! \Big|_{\cdot\dfrac{1}{(x-2)!}} \Rightarrow 6\cdot\dfrac{(x-5)!}{(x-7)!} \leq \dfrac{x!}{(x-2)!} \Rightarrow
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6\cdot\dfrac{(x-7)!(x-6)(x-5)}{(x-7)!}\leq \dfrac{(x-2)!(x-1)x}{(x-2)!}[/tex]
Simplificăm factorialele și rezultă:
[tex]\it 6(x-6)(x-5) \leq (x-1)x \Rightarrow 5x^2-65x+180 \leq0|_{:5} \Rightarrow
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x^2-13x+36\leq0 \ \ \ \ \ (2)
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x^2-13x+36 =0 \Rightarrow x_1=4,\ \ x_2=9 \ \ \ \ \ (3)
[/tex]
[tex]\it (2),\ (3) \Rightarrow x\in [4,\ 9] \ \ \ \ \ (4)
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(1),\ (4)\Rightarrow x\in \{7,\ 8,\ 9\}.[/tex]
