conditii de existenta: 15≥x
15∈N(A), x∈N
15≥x-2
x-2∈N
[tex] \frac{15!}{(15-x)!* x!} \leq \frac{15!}{(15-x+2)!*(x-2)!} => \frac{1}{(15-x)!*x!} \leq \frac{1}{(17-x)!*(x-2)!} [/tex] <-- am simplificat pe 15!
[tex] \frac{1}{(-x+15)!*x!} \leq \frac{1}{(-x+17)!*(x-2)!}[/tex] =>[tex] \frac{1}{(-x+15)!*(x-2)!(x-1)x} \leq \frac{1}{(-x+15)!(-x+16)(-x+17)(x-2)!}[/tex]
[tex] \frac{1}{(x-1)x} \leq \frac{1}{(16-x)(17-x)} => \frac{(16-x)(17-x)-(x-1)x}{(x-1)x(16-x)(17-x)} \leq 0[/tex]
=> 272-16x-17x+x²-(x²-x)=0 =>272-16x-17x+x²-x²+x=0 =>272-32x=0⇒x=272/32=17/2.
(x-1)(16-x)(17-x)x=0 ⇒ x=1,x=16,x=17,x=0
tabel de variatie:
x -inf 0 1 17/2 16 17 +inf
-32x+272 +++++++++++++0---------------------------
x-1 -----------------0++++++++++++++++++
16-x ++++++++++++++++0---------------------
17-x ++++++++++++++++++++0--------------
x -------0+++++++++++++++++++++++
toate parantezele inmultite ++++0-------0+++++++0------0+++++++
toata fractia ++++|-------|+++++0----|+++|--------------
=> x∈(0;1) reunit cu [17/2;16) reunit cu (17; inf)
verifici conditiile: 17/2 pica fiindca nu e nr natural; din intervale ⇒ solutie finala={9,10,...15} reunit cu {18,19,20,...,n,...}
