[tex] \frac{n-1}{n-3} - 20 \leq 0[/tex]
[tex] \frac{n-1}{n-3} - \frac{20(n-3)}{n-3} \leq 0[/tex]
n-1 - 20n + 60 totul supra n-3 ≤ 0
[tex] \frac{-19n+59}{n-3} [/tex]≤0
egalam numaratorul si numitorul cu 0.
-19n+59=0 ⇒-19n=-59⇒n= [tex] \frac{59}{19} [/tex]
n-3=0 ⇒ n=3
facem tabel de semne
n | -inf 3 [tex] \frac{59}{19} [/tex] +inf
-19n+59 |++++++++++++++++0----------------------------------------
n-3 |------------------0+++++++++++++++++++++++++++++
fractia |-------------------/+++++0----------------------------------------
solutie: n∈(-inf ; 3) reunit cu [ [tex] \frac{59}{19} [/tex]; +inf)
