3)
[tex]\it C_x^{x-2} = \dfrac{x!}{(x-2)!(x-x+2)!} = \dfrac{(x-2)!(x-1)x}{(x-2)! \cdot 2!}= \dfrac{(x-1)x}{2}[/tex]
[tex]\it A_x^3= \dfrac{x!}{(x-3)!} =\dfrac{(x-3)!(x-2)(x-1)x}{(x-3)!}=(x-2)(x-1)x[/tex]
Ecuaţia devine :
[tex]\it \dfrac{(x-1)x}{2} + (x-2)(x-1)x = 14x\ |:x \Longleftrightarrow [/tex]
[tex]\it \Longleftrightarrow \dfrac{x-1}{2}+(x-2)(x-1)=14\Longleftrightarrow (x-1) + 2(x-2)(x-1)=28 [/tex]
[tex]\it \Longleftrightarrow (x-1)(1+2x-4) = 28 \Longleftrightarrow (x-1)(2x-3)=28[/tex]
Pentru că x este număr natural, iar 4·7 = 28, va rezulta că x = 5
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