[tex]C_{5x}^{x^2+4} = 1 \\ \\ (1)~~ x^2+4 = 5x \Rightarrow x^2-5x+4 = 0 \Rightarrow (x-1)(x-4) = 0 \\\Rightarrow x_1 = 1,\quad x_2 = 4\\ \\ (2)~~x^2+4 = 0 \Rightarrow x^2 = -4 \Rightarrow x\in \emptyset \\ \\ \text{Nu mai facem conditiile de existenta, doar verificam, fiindca avem}\\ \text{doar 2 solutii posibile:}\\ \\ x = 1\Rightarrow C_{5}^{5} = 1 \quad (A) \\ x = 4 \Rightarrow C_{20}^{20} = 1~~(A) \\ \\ \Rightarrow x\in \{1;4\}[/tex]
