Răspunsuri

2014-10-01T19:09:06+03:00
 \frac{n-1}{n-3}  - 20  \leq 0
 \frac{n-1}{n-3} -  \frac{20(n-3)}{n-3}   \leq 0
n-1 - 20n + 60  totul supra n-3 ≤ 0
 \frac{-19n+59}{n-3} ≤0

egalam numaratorul si numitorul cu 0.
-19n+59=0 ⇒-19n=-59⇒n=  \frac{59}{19}
n-3=0 ⇒ n=3

facem tabel de semne
n            | -inf                3           \frac{59}{19}         +inf
-19n+59 |++++++++++++++++0----------------------------------------
n-3         |------------------0+++++++++++++++++++++++++++++
fractia    |-------------------/+++++0----------------------------------------

solutie:  n∈(-inf ; 3) reunit cu [  \frac{59}{19} ; +inf)

2014-10-02T00:43:27+03:00
Calculam produsul mezilor egal cu alextremilor si rezulta
n-1≤20(n-3)
n-1≤20n-60
60-1≤20n-n
59≤19n
59/19≤n
3,105≤n