Question

$A_x^7+3A_x^5=45A_x^5.$
 x=? Raspunsuri: x=8; x=0; x=12; x=13; x=7

Answer 1

Ax,7(nu am cum sa redactez altfel)=x!/(x-7)!
Ax,5=x!/(x-5)!
inlocuind
x!/(x-7)! =42x!/(x-5)!|:x!
(x-5)!/(x-7)!=42
(x-5)(x-6)(x-7)!/(x-7)!=42
(x-5)(x-6)=42
x²-11x+30-42=0
x²-11x-12=0

cu Δ=13
x1=-1
x2=12
S:x=12

Answer 2

$\it A^7_x +3A^5_x = 45 A^5_x \Leftrightarrow A^7_x = 45 A^5_x - 3A^5_x \Leftrightarrow A^7_x = 42 A^5_x \Leftrightarrow$

$\it \Leftrightarrow (x-6)(x-5) A^5_x =42A^5_x \Leftrightarrow (x-6)(x-5) =42\ \ \ \ (1)$

$\it x\in \mathbb{N},\ \ 6\cdot7 = 42 \ \ \ \ (2)$

$\it (1), (2) \Longrightarrow x-6=6 \Longrightarrow x=12\ .$