Question

am nevoie de ajutor, de la b în jos. mulțumesc. coroană.

Answer 1

   
[tex]\displaystyle\\ Ex. ~6)\\\\ b)\\ A_{x+2}^2 =56(x+2) \\ (x+2)(x+1) = 56(x+2) ~~~| :(x+2)\\ x+1 = 56\\ x = 56 - 1 = \boxed{55}\\\\ c)\\ A_{x+1}^2 =30 \\ (x+1)\cdot x=30\\ x(x+1)=30\\ \text{x si (x+1) sunt numere naturale consecutive.}\\ \text{Descompunem pe 30 in produs de numere consecutive.}\\ x(x+1)=5\cdot 6\\ \Longrightarrow~~~ x = \boxed{5} [/tex]


[tex]\displaystyle\\ d)\\ A_{n-1}^5 =18\cdot A_{n-3}^4 \\ (n-1)(n-2)(n-3)(n-4)(n-5) = 18(n-3)(n-4)(n-5)(n-6)\\ \text{Impartim ecuatia la: }(n-3)(n-4)(n-5) \\ (n-1)(n-2) = 18(n-6)\\ n^2 -3n +2 = 18n - 108\\ n^2 -3n-18n + 2+108 = 0\\ n^2 -21n +110 =0\\ n^2 -10n - 11n +110 = 0\\ n(n-10) -11(n-10)= 0\\ (n-10)(n-11)=0\\ n_1 = \boxed{10}\\ n_2 = \boxed{11}[/tex]


[tex]\displaystyle\\ e)\\ A_x^2+2A_{x+1}^2=30\\ x(x-1) + 2(x+1)\cdot x = 30\\ x(x-1) + 2x(x+1) = 30\\ x^2-x + 2x^2 + 2x=30\\ 3x^2 +x = 30\\ 3x^2 +x - 30=0\\ 3x^2 -9x + 9x +x-30 = 0\\ 3x^2 -9x + 10x -30 = 0\\ 3x(x-3)+ 10(x-3)=0\\ (x-3)(3x+10)=0\\ x-3 = 0 ~~~\Longrightarrow ~~~x = \boxed{3}\\ 3x+10=0~~~\Longrightarrow ~~~x \notin N[/tex]


[tex]\displaystyle\\ f)\\ A_{x+1}^{x-1}=360\\\\ \frac{P_{x+1}}{P_2}=360\\\\ \frac{(x+1)!}{2}=360\\\\ (x+1)! = 360 \cdot 2\\ (x+1)! = 720\\ (x+1)! = 6!\\ x+1 = 6\\ x = 6-1 = \boxed{5}[/tex]


[tex]\displaystyle\\ g)\\ A_{x+3}^{x+2}=10A_{x+2}^{x+1}\\\\ \frac{P_{x+3}}{P_1} = 10\cdot \frac{P_{x+2}}{P_1}\\\\ \frac{(x+3)!}{1} = 10\cdot \frac{(x+2)!}{1} \\\\ (x+3)! = 10\cdot (x+2)!\\\\ \frac{(x+3)!}{(x+2)!} =10\\\\ \frac{(x+2)! \cdot (x+3)}{(x+2)!} =10\\\\ x+3 = 10 x = 10 - 3 = \boxed{7}[/tex]