Question

ex 10 punctul a) si b)
va rog !!!!

Answer 1

a) 

$\dfrac{x!}{5!(x-5)!}>\dfrac{x!}{6!(x-6)!}\\ \\ \\ 5!(x-6)!\cdot (x-5)<5! \cdot 6 \cdot (x-6)!\\ \\ \\ x-5<6\\ \\ \\ x<11.$

Answer 2

 10a)
$C_x^{5}>C_x^{6} \\ \\ \frac{x(x-1)(x-2)(x-3)(x-4)}{5!} > \frac{x(x-1)(x-2)(x-3)(x-4)(x-5)}{6!} \\ \\ \frac{6!}{5!} > \frac{x(x-1)(x-2)(x-3)(x-4)(x-5)}{x(x-1)(x-2)(x-3)(x-4)} \\ \\ 6 > x-5 \\ x<6+5 \\ x<11 \\ Dar\,\,x\,\,nu\,poate\,fi\,mai\,mic\,decat\,\,6. \\ => \boxed{ 6 \leq x<11}$
  
10b)   
$C_{x-1}^4-C_{x-1}^3 - \frac{5*A_{x-2}^2}{4} <0 \\ \\ \frac{(x-1)(x-2)(x-3)(x-4)}{4!} - \frac{(x-1)(x-2)(x-3)}{3!}-\frac{5*(x-2)(x-1)}{4} <0 \\ \\ \frac{(x-1)(x-2)(x-3)(x-4)}{4!}- \frac{4(x-1)(x-2)(x-3)}{4!}-\frac{3!*5*(x-2)(x-1)}{4!} <0 \\ \\ \frac{(x-1)(x-2)(x-3)(x-4)-4(x-1)(x-2)(x-3)-3!*5(x-2)(x-1)}{4!}<0 \\ (x-1)(x-2)[(x-3)(x-4)-4(x-3)-6*5] <0 \\ (x-3)(x-4)-4(x-3)-6*5<0 \\ x^{2} -7x+12-4x+12-30<0 \\ x^{2} -11x-6 <0 \\ x_1 = (11 + \sqrt{121+4*6})/2 = (11 + \sqrt{145})/2 =aprox11,5$

$x_2 = (11 - \sqrt{121+4*6})/2 = (11 - \sqrt{145})/2 =aprox-0,5 \\ Coeficientul \,\,lui\,\, x^{2} \,\,este\,\, pozitiv \\ =>\boxed{ 5 \leq x \leq 11}$