+FML Posted April 26, 2015 Report Posted April 26, 2015 Se da inecuatia : '' Aranjamente de x , luate cate x - 1 '' + '' Aranjamente de x , luate cate x - 2 '' <= 5(n-1)! Primul aranjament vine : x!/(x-x-1)! = x!/-1! [ e negativ ] . Al doilea : x!/(x-x-2)! = x!/-2! [ tot negativ ] . Ei , ce fac acum ?
Moderators Courage Posted April 26, 2015 Moderators Report Posted April 26, 2015 Primul aranjament vine : x!/(x-x-1)! = x!/-1! [ e negativ ] .Adica x! / (x-(x-1))! = x! / (x-x+1)! = x! / 1! = x!
+FML Posted April 26, 2015 Author Report Posted April 26, 2015 Adica x! / (x-(x-1))! = x! / (x-x+1)! = x! / 1! = x! Ahh , uitasi paranteza . Pai si cu 5(n-1)! ce fac ?
Moderators Courage Posted April 26, 2015 Moderators Report Posted April 26, 2015 Poate 5(x-1)!, altfel nu vad rostul.
+FML Posted April 26, 2015 Author Report Posted April 26, 2015 Poate 5(x-1)!, altfel nu vad rostul. Nu mai stiu sigur , dar parca da . Si ce - i fac ?
Moderators Courage Posted April 26, 2015 Moderators Report Posted April 26, 2015 Pai te folosesti de faptul ca (x-1)! = x! / x
+FML Posted April 26, 2015 Author Report Posted April 26, 2015 Pai te folosesti de faptul ca (x-1)! = x! / x Deci , de asemenea , o sa am 5 (x!/x) = 5x!/5x , nu ?
+FML Posted April 26, 2015 Author Report Posted April 26, 2015 5 * x! / x Pai , paranteza inseamna tot ori . Nu e bine daca pun alea in paranteza ?
Moderators Courage Posted April 26, 2015 Moderators Report Posted April 26, 2015 5*(a/b) nu inseamna 5a/5b.
+FML Posted April 26, 2015 Author Report Posted April 26, 2015 5*(a/b) nu inseamna 5a/5b. cum nai'... Pai nu se inmulteste numarul din fata cu fiecare termen al parantezei ? Sau aia era doar pentru anumite operatii ?
+FML Posted April 26, 2015 Author Report Posted April 26, 2015 ahh , da , eu incepusem sa ma gandesc la inmultime ( a*b ) . Bun . Deci avem : x!+(x!/2) = 5x!/x Nu ?
+FML Posted April 26, 2015 Author Report Posted April 26, 2015 Mersi ! Dar nu pricep ultimele 2 . La mine da 2/2 ori x .
+FML Posted April 26, 2015 Author Report Posted April 26, 2015 Stiu asta , da' nu stiu de unde - l iei tu pe al treilea . ah , stai ma ca se amplifica primul ''a'' ca sa fie supra 2 , si vine 2a/2 + a /2 = 3a/2 . Scuze , nu stiu ce am . In continuare stiu ce trebuie . Mersi ca m - ai suportat . Cerebelul meu intra - n vacanta [ iar ] .
Moderators Courage Posted April 26, 2015 Moderators Report Posted April 26, 2015 Inmultesti cu 1/x! si da
+FML Posted April 26, 2015 Author Report Posted April 26, 2015 Inmultesti cu 1/x! si da Corect , dar nu stiu sa - i explic profului de ce as face asta . P.S : Cumva ca sa scap de factorial ? LE : Chiar si - asa , nu vad cum dispar alea ...
Moderators Courage Posted April 26, 2015 Moderators Report Posted April 26, 2015 x! nu va fi niciodata 0, deci poti inmulti cu 1/x! ca sa scapi de el din ambii membri.
+FML Posted April 26, 2015 Author Report Posted April 26, 2015 x! nu va fi niciodata 0, deci poti inmulti cu 1/x! ca sa scapi de el din ambii membri. 0! = 1 [ coventie ] . Multam fain , daca te prind iti fac cinste , promit . PS : Finally 3x<= 10 x<= 10/7 x (= [ apartine ] ( -00 , 1o/7 ] .
+FML Posted April 27, 2015 Author Report Posted April 27, 2015 Da , produsul mezilor = produsul extremilor .
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