Question

Aranjamente

Să se rezolve ecuațiile:

n (jos) A k (sus)

a) x-2 A 2 = 42
b) x+2 A 2 = 56 x (x+2)
c) x+1 A 2 = 30
d) n-1 A 5 = 18 x n-3 A 4
e) x A 2 + 2 x+1 A 2 = 30
f) x+1 A x-1 = 360
g) x+3 A x+2 = 10 x+2 A x + 1

Answer 1

$\displaystyle a).A_{x-2}^2=42~~~~~~~~~~~~~~~~~~x-2 \geq 2 \Rightarrow x \geq 4 \Rightarrow x \in \{4,5,6,...\}=D \\ \\ \frac{(x-2)!}{(x-2-2)!} =42 \Rightarrow \frac{(x-2)!}{(x-4)!} =42 \Rightarrow (x-2)(x-3)=42 \Rightarrow \\ \\ \Rightarrow x^2-3x-2x+6=42 \Rightarrow x^2-5x+6-42=0 \Rightarrow \\ \\ \Rightarrow x^2-5x-36=0 \\ \\ a=1,~b=-5,~c=-36 \\ \\ \Delta=b^2-4ac=(-5)^2-4 \cdot 1 \cdot (-36)=25+144=169\ \textgreater \ 0 \\ \\ x_1= \frac{5+ \sqrt{169} }{2 \cdot 1} = \frac{5+13}{2} = \frac{18}{2} =9$

$\displaystyle x_2= \frac{5- \sqrt{169} }{2 \cdot 1} = \frac{5-13}{2} = \frac{-8}{2} =-4 \\ \\ S=\{9\}$

$\displaystyle b).A_{x+2}^2=56 \cdot (x+2)~~~~~~~~~~x+2 \geq 2 \Rightarrow x \geq 0 \Rightarrow x \in \{0,1,2,...\}=D \\ \\ \frac{(x+2)!}{(x+2-2)!} =56x+112 \Rightarrow \frac{(x+2)!}{x!} =56x+112 \Rightarrow \\ \\ \Rightarrow(x+2)(x+1)=56x+112 \Rightarrow x^2+x+2x+2=56x+112 \Rightarrow \\ \\ \Rightarrow x^2+3x-56x=112-2 \Rightarrow x^2-53x=110 \Rightarrow x^2-53x-110=0 \\ \\ a=1,~b=-53,~c=-110 \\ \\ \Delta=b^2-4ac=(-53)^2-4 \cdot 1 \cdot (-110)=2809+440=3249\ \textgreater \ 0$

$\displaystyle x_1= \frac{53+ \sqrt{3249} }{2 \cdot 1} = \frac{53+57}{2} = \frac{110}{2} =55 \\ \\ x_2= \frac{53- \sqrt{3249} }{2 \cdot 1} = \frac{53-57}{2} = \frac{-4}{2} =-2 \\ \\ S=\{55\}$

$\displaystyle c).A_{x+1}^2=30~~~~~~~~~~~~~~~x+1 \geq 2 \Rightarrow x \geq 1 \Rightarrow x \in \{1,2,3,...\}=D \\ \\ \frac{(x+1)!}{(x+1-2)!} =30 \Rightarrow \frac{(x+1)!}{(x-1)!} =30 \Rightarrow x(x+1)=30 \Rightarrow \\ \\ \Rightarrow x^2+x=30 \Rightarrow x^2+x-30=0 \\ \\ a=1,~b=1,~c=-30\\ \\ \Delta=b^2-4ac=1^2-4 \cdot 1 \cdot (-30)=1+120=121\ \textgreater \ 0$

$\displaystyle x_1=\frac{-1+ \sqrt{121} }{2 \cdot 1}=\frac{-1+11}{2} = \frac{10}{2}=5\\ \\x_2=\frac{-1-\sqrt{121} }{2 \cdot 1} =\frac{-1-11}{2} = \frac{-12}{2}=-6 \\ \\S=\{5\}$

$\displaystyle d).A_{n-1}^5=18 \cdot A_{n-3}^4 \left\begin{array}{ccc}&&n-1 \geq 5 \Rightarrow n \geq 6\\&&n-3 \geq 4 \Rightarrow n \geq 7\\\end{array}\right] \Rightarrow n \in \{7,8,9,...\}=D \\ \\ \frac{(n-1)!}{(n-1-5)!} =18 \cdot \frac{(n-3)!}{(n-3-4)!} \Rightarrow \frac{(n-1)!}{(n-6)!} =18 \cdot \frac{(n-3)!}{(n-7)!} \Rightarrow \\ \\ \Rightarrow (n-1)(n-2)(n-3)(n-4)(n-5)=18 \cdot(n-3)(n-4)(n-5)(n-6) \Rightarrow$

$\displaystyle \Rightarrow (n-3)(n-4)(n-5)(n-10)(n-11)=0 \\ \\ n-3=0 \Rightarrow n=3 \\ \\ n-4=0 \Rightarrow n=4 \\ \\ n-5=0 \Rightarrow n=5 \\ \\ n-10=0 \Rightarrow n=10 \\ \\ n-11=0 \Rightarrow n=11 \\ \\ S=\{10,11\}$

$\displaystyle e).A_x^2+2A_{x+1}^2=30~ \left\begin{array}{ccc}&&x \geq 2\\&&x+1 \geq 2\Rightarrow x \geq 1\\\end{array}\right] \Rightarrow x \in \{2,3,4,...\}=D \\ \\ \frac{x!}{(x-2)!} +2 \cdot \frac{(x+1)!}{(x+1-2)!}=30 \Rightarrow \frac{x!}{(x-2)!} +2 \cdot \frac{(x+1)!}{(x-1)!} =30 \Rightarrow \\ \\ \Rightarrow x(x-1)+2 \cdot x(x+1)=30 \Rightarrow x^2-x+2(x^2+x)=30 \Rightarrow \\ \\ \Rightarrow x^2-x+2x^2+2x=30 \Rightarrow 3x^2+x-30=0 \\ \\a=3,~b=1,~c=-30$

$\displaystyle \Delta=b^2-4ac=1^2-4 \cdot 3 \cdot(-30)=1+360=361\ \textgreater \ 0 \\ \\ x_1= \frac{-1+ \sqrt{361} }{2 \cdot 3} = \frac{-1+19}{6} = \frac{18}{6} =3 \\ \\ x_2= \frac{-1- \sqrt{361} }{2 \cdot 3} = \frac{-1-19}{6} =- \frac{20}{6} =- \frac{10}{3} \\ \\ S=\{3\}$

$\displaystyle f). A_{x+1}^{x-1}=360 \Rightarrow \frac{(x+1)!}{(x+1-x+1)!} =360 \Rightarrow \frac{(x+1)!}{2!} =360 \Rightarrow \\ \\ \Rightarrow \frac{(x+1)!}{1 \cdot 2} =360 \Rightarrow \frac{(x+1)!}{2} =360 \Rightarrow (x+1)!=360 \cdot2 \Rightarrow \\ \\ \Rightarrow (x+1)!=720 \Rightarrow x+1=6 \Rightarrow x=6-1 \Rightarrow x=5 \\ \\ S=\{5\}$

$\displaystyle g).A_{x+3}^{x+2}=10A_{x+2}^{x+1} \Rightarrow \frac{(x+3)!}{(x+3-x-2)!} =10 \cdot \frac{(x+2)!}{(x+2-x-1)!} \Rightarrow \\ \\ \Rightarrow \frac{(x+3)!}{1!} =10 \cdot \frac{(x+2)!}{1!} \Rightarrow (x+3)!=10 \cdot (x+2)! \Rightarrow \\ \\ \Rightarrow \frac{(x+3)!}{(x+2)!} =10 \Rightarrow x+3=10 \Rightarrow x=10-3 \Rightarrow x=7 \\ \\ S=\{7\}$