Question

Combinări

Să se rezolve ecuațiile:

n (jos) C k (sus)

a) x-2 C 2 = 21
b) x+2 C 2 = 15
c) x+2 C 2 = 36
d) x C x-3 = 2 x C x-2
e) x C 3 = 5/4 x(x-3)
f) x+1 C 3 + x+1 C 2 = 15x
g) 5 x C 3 = x+2 C 4
h) 12 1 C x + x+4 C 2 = 162

Answer 1

[tex]\displaystyle a).C_{x-2}^2=21 ~~~~~~~~~~~~~~~~~x-2 \geq 2 \Rightarrow x \geq 4 \Rightarrow x \in \{4,5,6,...\}=D \\ \\ \frac{(x-2)!}{(x-2-2)!\cdot2!} =21 \Rightarrow \frac{(x-2)!}{(x-4)!\cdot 2!} =21 \Rightarrow \frac{(x-2)(x-3)}{2!} =21 \Rightarrow \\ \\ \Rightarrow \frac{(x-2)(x-3)}{1 \cdot 2} =21 \Rightarrow \frac{(x-2)(x-3)}{2} =21 \Rightarrow \\ \\ \Rightarrow (x-2)(x-3)=21 \cdot 2 \Rightarrow x^2-3x-2x+6=42 \Rightarrow \\ \\ \Rightarrow x^2-5x+6-42=0 \Rightarrow x^2-5x-36=0 [/tex]

[tex]\displaystyle 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 \\ \\ x_2= \frac{5- \sqrt{169} }{2 \cdot 1} = \frac{5-13}{2} = \frac{-8}{2} =-4 \\ \\ S=\{9\}[/tex]

$\displaystyle b).C_{x+2}^2=15~~~~~~~~~~~~~~~~~~~x+2 \geq 2 \Rightarrow x \geq 0 \Rightarrow x \in \{0,1,2,...\}=D \\ \\ \frac{(x+2)!}{(x+2-2)! \cdot 2!} =15 \Rightarrow \frac{(x+2)!}{x! \cdot 2!} =15 \Rightarrow \frac{(x+2)!}{2x!} =15 \Rightarrow \\ \\ \Rightarrow \frac{(x+2)(x+1)}{2} =15 \Rightarrow (x+2)(x+1)=15 \cdot 2 \Rightarrow \\ \\ \Rightarrow x^2+x+2x+2=30 \Rightarrow x^2+3x+2-30=0 \Rightarrow \\ \\ \Rightarrow x^2+3x-28=0$

[tex]\displaystyle a=1,~b=3,~c=-28 \\ \\ \Delta=b^2-4ac=3^2-4 \cdot 1 \cdot (-28)=9+112=121\ \textgreater \ 0 \\ \\ x_1= \frac{-3+ \sqrt{121} }{2 \cdot 1} = \frac{-3+11}{2} = \frac{8}{2} =4 \\ \\ x_2= \frac{-3- \sqrt{121} }{2 \cdot 1} = \frac{-3-11}{2} = \frac{-14}{2} =-7 \\ \\ \\ S=\{4\}[/tex]

$\displaystyle c).C_{x+2}^2=36~~~~~~~~~~~~~~~~~~~~x+2 \geq 2 \Rightarrow x \geq 4 \Rightarrow x \in \{4,5,6,...\}=D \\ \\ \frac{(x+2)!}{(x+2-2)! \cdot 2!} =36 \Rightarrow \frac{(x+2)!}{x! \cdot 2!}=36 \Rightarrow \frac{(x+2)!}{2x!} =36 \Rightarrow \\ \\ \Rightarrow \frac{(x+2)(x+1)}{2} =36 \Rightarrow (x+2)(x+1)=36 \cdot 2\Rightarrow \\ \\ \Rightarrow x^2+x+2x+2=72 \Rightarrow x^2+2x+2-72=0 \Rightarrow x^2+3x-70=0$

[tex]\displaystyle a=1,~b=3,~c=-70 \\ \\ \Delta=b^2-4ac=3^2-4 \cdot 1 \cdot (-70)=9+280=289\ \textgreater \ 0 \\ \\ x_1= \frac{-3+ \sqrt{289} }{2 \cdot 1} = \frac{-3+17}{2} = \frac{14}{2} =7 \\ \\ x_2= \frac{-3- \sqrt{289} }{2 \cdot 1} = \frac{-3-17}{2} = \frac{-20}{2} =-10 \\ \\ S=\{7\}[/tex]

$\displaystyle d).C_x^{x-3}=2C_x^{x-2} \Rightarrow \frac{x!}{(x-x+3)!\cdot (x-3)!} =2 \cdot \frac{x!}{(x-x+2)! \cdot (x-2)!} \Rightarrow \\ \\ \Rightarrow \frac{x!}{3! \cdot (x-3)!} = \frac{2x!}{2! \cdot (x-2)!} \Rightarrow \frac{x(x-1)(x-2)}{3!} = \frac{2x(x-1)}{2!} \Rightarrow \\ \\ \Rightarrow \frac{x(x-1)(x-2)}{1 \cdot 2 \cdot 3} = \frac{2x(x-1)}{1 \cdot 2} \Rightarrow \frac{x(x-1)(x-2)}{6} = \frac{2x(x-1)}{2} \Rightarrow$

$\displaystyle \Rightarrow \frac{x(x-1)(x-2)}{6} =x(x-1) \Rightarrow \frac{(x^2-x)(x-2)}{6} =x^2-x \Rightarrow \\ \\ \Rightarrow (x^2-x)(x-2)=6(x^2-x) \Rightarrow x^3-2x^2-x^2+2x=6x^2-6x \Rightarrow \\ \\ \Rightarrow x^3-3x^2+2x=6x^2-6x \Rightarrow x^3-3x^2+2x-6x^2+6x=0 \Rightarrow \\ \\ x^3-9x^2+8x=0 \\ \\ x(x-1)(x-8)=0 \\ \\ x=0 \\ \\ x-1=0 \Rightarrow x=1 \\ \\ x-8=0 \Rightarrow x=8 \\ \\ S=\{8\}$

$\displaystyle e).C_x^3= \frac{5}{4} x(x-3)~~~~~~~~~~~~~~~~~~~~~~~~~x \geq 3 \Rightarrow x \in \{3,4,5,...\}=D \\ \\ \frac{x!}{(x-3)!\cdot 3!} = \frac{5}{4} (x^2-3x) \Rightarrow \frac{x(x-1)(x-2)}{3!} = \frac{5}{4} (x^2-3x) \Rightarrow \\ \\ \Rightarrow \frac{(x^2-x)(x-2)}{1 \cdot 2 \cdot 3} = \frac{5x^2}{4} - \frac{15x}{4} \Rightarrow \frac{(x^2-x)(x-2)}{6} = \frac{5x^2-15x}{4} \Rightarrow$

$\displaystyle \Rightarrow \frac{x^3-2x^2-x^2+2x}{6} = \frac{5x^2-15x}{4} \Rightarrow \frac{x^2-3x^2+2x}{6} = \frac{5x^2-15x}{4} \Rightarrow \\ \\ \Rightarrow 4(x^3-3x^2+2x)=6(5x^2-15x) \Rightarrow \\ \\ \Rightarrow 4x^3-12x^2+8x=30x^2-90x \Rightarrow \\ \\ \Rightarrow 4x^3-12x^2+8x-30x^2+90x=0 \Rightarrow 4x^3-42x^2+98x=0 \Rightarrow \\ \\ \Rightarrow 2x(x-7)(2x-7)=0 \\ \\ x=0 \\ \\ x-7=0 \Rightarrow x=7 \\ \\ 2x-7=0 \Rightarrow x= \frac{7}{2} \\ \\ S=\left\{ \frac{7}{2} ,7\right\}$