Versiune completa: Mate Cls A V A !
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Buna seara sunt mamica a unui baietel de cls a v a care ma dispera cu pb la matematica;Nu a inteles niciodata Teorema impartirii cu rest sau mai degraba in aplicatii ,nu stiu ;cert este ca uneorma depasesc si pe mine pb ptr ca nu prea gasesc ex rezolvate in caiet ca sa ma pot orienta si apoi sa i le pot explica;AJUTA-TI MA ,va rog!
daca pot , te ajut cu ce-a mai mare placere , eu sunt in clasa a 8a si mai stiu cate ceva 
QUOTE(98brv @ Oct 1 2009, 10:23 PM) 
Buna seara sunt mamica a unui baietel de cls a v a care ma dispera cu pb la matematica;Nu a inteles niciodata Teorema impartirii cu rest sau mai degraba in aplicatii ,nu stiu ;cert este ca uneorma depasesc si pe mine pb ptr ca nu prea gasesc ex rezolvate in caiet ca sa ma pot orienta si apoi sa i le pot explica;AJUTA-TI MA ,va rog!
Si noi tot in clasa a cincea suntem si ne pricepem binisor la mate....daca te putem ajuta....
buna,
noi tot a cincea sintem,da-ne un exemplu pe care vrei ca noi sa ti-l explicam!noi am facut impartirea cu rest in clasa a 4 a ,cred ca si la voi tot asa a fost ,nu?eventual sa te uiti si in manual ,stiu ca pe vremea mea se dadea si un exemplu cum sa face in carte!
te inteleg perfect ,scriu astia niste texte ,la problemele de logica (baiatul meu a facut clasa 1 si 2 in Ro)de sincer trebuia sa citesc de 2 ori enuntul problemei,eu ca adult ,darmite un copil...nu stiu de ce se complica atit!
aici e totul simplist si se invata si usor asa!pai nemtii deja facusera tabla inmulturii in clasa a doua ,iar baiatul meu cind a venit aici nici nu auzise de ea ,se invata in a treia in ro,dar era as in probleme de logica cu necunoscute ...care aici nu i-au folosit la nimic!
noi acum sintem la numere de ordinul mil,miliardelor ...bilioanelor ,biliardelor...si la sistemul decimat,voi ce studiati acum la mate?
noi tot a cincea sintem,da-ne un exemplu pe care vrei ca noi sa ti-l explicam!noi am facut impartirea cu rest in clasa a 4 a ,cred ca si la voi tot asa a fost ,nu?eventual sa te uiti si in manual ,stiu ca pe vremea mea se dadea si un exemplu cum sa face in carte!
te inteleg perfect ,scriu astia niste texte ,la problemele de logica (baiatul meu a facut clasa 1 si 2 in Ro)de sincer trebuia sa citesc de 2 ori enuntul problemei,eu ca adult ,darmite un copil...nu stiu de ce se complica atit!
aici e totul simplist si se invata si usor asa!pai nemtii deja facusera tabla inmulturii in clasa a doua ,iar baiatul meu cind a venit aici nici nu auzise de ea ,se invata in a treia in ro,dar era as in probleme de logica cu necunoscute ...care aici nu i-au folosit la nimic!
noi acum sintem la numere de ordinul mil,miliardelor ...bilioanelor ,biliardelor...si la sistemul decimat,voi ce studiati acum la mate?
multumesc ptr promptitudine;ei nu au facut in cls a IV a teorema imp cu rest ,abia acu in clasa a Va ;oricum sa zicem ca"a inteles-o";trec mai departe la alte lectii ;in caietul de clasa au ceva ,tema este despre altceva nu stiu pe mine ma cam depasesc uneori;daca as avea un exemplu pe caietul de clasa......m-as lamuri;in fine acum au ca tema printre altele si niste exercitii la care mi-am prins urechile;daca ma puteti ajuta......iata:
Daca x+2y=17 si y-3z=4 ,calculati:
a) x+3y-3z;
b)2x+5y-3z;
c)(3x+6y)+(2y-6z):
d)(2x+4y) X(x+4y-6z);( X este semnul inmultirii)
macar un exemplu daca mi-l explicati poate mai departe i le explic eu ,le facem impreuna..Nu prea ne place matematica
Multumesc
Daca x+2y=17 si y-3z=4 ,calculati:
a) x+3y-3z;
b)2x+5y-3z;
c)(3x+6y)+(2y-6z):
d)(2x+4y) X(x+4y-6z);( X este semnul inmultirii)
macar un exemplu daca mi-l explicati poate mai departe i le explic eu ,le facem impreuna..Nu prea ne place matematica
Multumesc
QUOTE(98brv @ Oct 4 2009, 01:55 PM)
multumesc ptr promptitudine;ei nu au facut in cls a IV a teorema imp cu rest ,abia acu in clasa a Va ;oricum sa zicem ca"a inteles-o";trec mai departe la alte lectii ;in caietul de clasa au ceva ,tema este despre altceva nu stiu pe mine ma cam depasesc uneori;daca as avea un exemplu pe caietul de clasa......m-as lamuri;in fine acum au ca tema printre altele si niste exercitii la care mi-am prins urechile;daca ma puteti ajuta......iata:
Daca x+2y=17 si y-3z=4 ,calculati:
a) x+3y-3z;
b)2x+5y-3z;
c)(3x+6y)+(2y-6z):
d)(2x+4y) X(x+4y-6z);( X este semnul inmultirii)
macar un exemplu daca mi-l explicati poate mai departe i le explic eu ,le facem impreuna..Nu prea ne place matematica
Multumesc
Daca x+2y=17 si y-3z=4 ,calculati:
a) x+3y-3z;
b)2x+5y-3z;
c)(3x+6y)+(2y-6z):
d)(2x+4y) X(x+4y-6z);( X este semnul inmultirii)
macar un exemplu daca mi-l explicati poate mai departe i le explic eu ,le facem impreuna..Nu prea ne place matematica
Multumesc
Se dau:
x+2y=17
y-3z=4
a )
Adunam, membru cu membru ecuatiile date initial:
(x+2y)+(y-3z)=17+4
x+2y+y-3z=21
x+3y-3z=21
b )
2*(x+2y)+(y-3z)=2*17+4
2x+4y+y-3z=38
2x+5y-3z=38
c )
3*(x+2y)+2*(y-3z)=3*17+2*4
(3x+6y)+(2y-6z)=51+8=59
d) Incercati!?...
Am folosit * pentru semnul inmultirii
QUOTE(Matei @ Oct 4 2009, 12:26 PM)
Se dau:
x+2y=17
y-3z=4
a )
Adunam, membru cu membru ecuatiile date initial:
(x+2y)+(y-3z)=17+4
x+2y+y-3z=21
x+3y-3z=21
b )
2*(x+2y)+(y-3z)=2*17+4
2x+4y+y-3z=38
2x+5y-3z=38
c )
3*(x+2y)+2*(y-3z)=3*17+2*4
(3x+6y)+(2y-6z)=51+8=59
d) Incercati!?...
Am folosit * pentru semnul inmultirii
x+2y=17
y-3z=4
a )
Adunam, membru cu membru ecuatiile date initial:
(x+2y)+(y-3z)=17+4
x+2y+y-3z=21
x+3y-3z=21
b )
2*(x+2y)+(y-3z)=2*17+4
2x+4y+y-3z=38
2x+5y-3z=38
c )
3*(x+2y)+2*(y-3z)=3*17+2*4
(3x+6y)+(2y-6z)=51+8=59
d) Incercati!?...
Am folosit * pentru semnul inmultirii
multumim ptr rezolvari ;am inteles metoda dar la ultima ne-am impotmolit;multumim ffff mult;avem cam multe "goluri" ...
buna Matei
am cateva exercitii de cl a
am cateva exercitii de cl a
BUNA MATEI
AM CATEVA EXERCITII DE CL A V-A PE CARE NU LE POT REZOLVA
1.ARATATI CA NR abcabc SE IMPARTE EXACT PRIN 1001
2.DETERMINATI NR NATURALE DE DOUA CIFRE SCRISE IN BAZA ZECE DE FORMA ab pt care ab=4(a+
.
AM CATEVA EXERCITII DE CL A V-A PE CARE NU LE POT REZOLVA
1.ARATATI CA NR abcabc SE IMPARTE EXACT PRIN 1001
2.DETERMINATI NR NATURALE DE DOUA CIFRE SCRISE IN BAZA ZECE DE FORMA ab pt care ab=4(a+
.
QUOTE(laura77 @ Oct 29 2009, 08:45 AM)
BUNA MATEI
AM CATEVA EXERCITII DE CL A V-A PE CARE NU LE POT REZOLVA
1.ARATATI CA NR abcabc SE IMPARTE EXACT PRIN 1001
2.DETERMINATI NR NATURALE DE DOUA CIFRE SCRISE IN BAZA ZECE DE FORMA ab pt care ab=4(a+.
AM CATEVA EXERCITII DE CL A V-A PE CARE NU LE POT REZOLVA
1.ARATATI CA NR abcabc SE IMPARTE EXACT PRIN 1001
2.DETERMINATI NR NATURALE DE DOUA CIFRE SCRISE IN BAZA ZECE DE FORMA ab pt care ab=4(a+
.Solutie:
1
abcabc=100000a+10000b+1000c++100a+10b+c=100100a+10010b+1001c=1001*100a+1001*10b+
001c=
=1001*(100a+10b+c) ...ceea ce terbuia demonstrat
2
10a+b=4a+4b
6a=3b (:3)
2a=b
De aici rezulta posibilitatile:
12=4(1+2)
24=4(2+4)
36=4(3+6)
48=4(4+8)
Adica numerele sunt: 12, 24, 36, 48
QUOTE(Matei @ Oct 29 2009, 09:31 AM)
Solutie:
1
abcabc=100000a+10000b+1000c++100a+10b+c=100100a+10010b+1001c=1001*100a+1001*10b+
001c=
=1001*(100a+10b+c) ...ceea ce terbuia demonstrat
2
10a+b=4a+4b
6a=3b (:3)
2a=b
De aici rezulta posibilitatile:
12=4(1+2)
24=4(2+4)
36=4(3+6)
48=4(4+8)
Adica numerele sunt: 12, 24, 36, 48
1
abcabc=100000a+10000b+1000c++100a+10b+c=100100a+10010b+1001c=1001*100a+1001*10b+
001c=
=1001*(100a+10b+c) ...ceea ce terbuia demonstrat
2
10a+b=4a+4b
6a=3b (:3)
2a=b
De aici rezulta posibilitatile:
12=4(1+2)
24=4(2+4)
36=4(3+6)
48=4(4+8)
Adica numerele sunt: 12, 24, 36, 48
multumesc mult
te mai deranjez cu 2ex
verificati daca urmatoarele egal. sunt adev
a. aa=11a
b. abab=101ab
c.a00a=1001a
2.determ nr nat consecutive ab si ac pt care ab +ac=113
QUOTE(laura77 @ Oct 29 2009, 09:42 AM)
multumesc mult
te mai deranjez cu 2ex
verificati daca urmatoarele egal. sunt adev
a. aa=11a
b. abab=101ab
c.a00a=1001a
2.determ nr nat consecutive ab si ac pt care ab +ac=113
te mai deranjez cu 2ex
verificati daca urmatoarele egal. sunt adev
a. aa=11a
b. abab=101ab
c.a00a=1001a
2.determ nr nat consecutive ab si ac pt care ab +ac=113
Solutie:
1 )La fel ca la ex. anterior:
a ) 10a+a=11a
b ) 1000a+100b+10a+b=1010a+101b=101(10a+b )=101ab
c ) 1000a+100*0+10*0+a=1001a
2 )
ab= 10a+b
ac= 10a+b+1 (b si c sunt consecutive...)
Deci,
10a+b+10a+b+1=113
20a+2b=112 (:2)
10a+b=56
Deci numerele sunt 56 si 57
Observatie: marea filozofie a acestor exercitii este scrierea numerelor in sistemul zecimal folosind puterile lui 10
exemplu:
abcd= (10**3)a+(10**2)b+10c+d
unde, spre exemplu, 10**3= 10*10*10=1000
QUOTE(Matei @ Oct 29 2009, 10:26 AM)
Solutie:
1 )La fel ca la ex. anterior:
a ) 10a+a=11a
b ) 1000a+100b+10a+b=1010a+101b=101(10a+b )=101ab
c ) 1000a+100*0+10*0+a=1001a
2 )
ab= 10a+b
ac= 10a+b+1 (b si c sunt consecutive...)
Deci,
10a+b+10a+b+1=113
20a+2b=112 (:2)
10a+b=56
Deci numerele sunt 56 si 57
Observatie: marea filozofie a acestor exercitii este scrierea numerelor in sistemul zecimal folosind puterile lui 10
exemplu:
abcd= (10**3)a+(10**2)b+10c+d
unde, spre exemplu, 10**3= 10*10*10=1000
1 )La fel ca la ex. anterior:
a ) 10a+a=11a
b ) 1000a+100b+10a+b=1010a+101b=101(10a+b )=101ab
c ) 1000a+100*0+10*0+a=1001a
2 )
ab= 10a+b
ac= 10a+b+1 (b si c sunt consecutive...)
Deci,
10a+b+10a+b+1=113
20a+2b=112 (:2)
10a+b=56
Deci numerele sunt 56 si 57
Observatie: marea filozofie a acestor exercitii este scrierea numerelor in sistemul zecimal folosind puterile lui 10
exemplu:
abcd= (10**3)a+(10**2)b+10c+d
unde, spre exemplu, 10**3= 10*10*10=1000
acum am inteles multumesc mult
p.s.
cred ca ex. 2 e gresit in carte
daca 2a=b nu avem sol.
12,24....,fiindca nu e posibil
sol ar fi21,42,dar nu verifica egalitatea ab=4(a+
QUOTE(laura77 @ Oct 29 2009, 11:00 AM)
acum am inteles multumesc mult
p.s.
cred ca ex. 2 e gresit in carte
daca 2a=b nu avem sol.
12,24....,fiindca nu e posibil
sol ar fi21,42,dar nu verifica egalitatea ab=4(a+
p.s.
cred ca ex. 2 e gresit in carte
daca 2a=b nu avem sol.
12,24....,fiindca nu e posibil
sol ar fi21,42,dar nu verifica egalitatea ab=4(a+
scuze,e corect ex eu nu am fost atenta,multumesc
QUOTE(laura77 @ Oct 29 2009, 11:04 AM)
scuze,e corect ex eu nu am fost atenta,multumesc
buna seara
QUOTE(laura77 @ Nov 3 2009, 07:22 PM)
buna seara
te rog ajuta-ma la aceste ex
1.afla x daca
x :2006=5rest r
2.afla x daca
x :4=a rest b
x :1 0=b rest a
3. afla r daca
a=1 x 2x .....x18 +2006
a:125=c rest r
4.x=?
x=84a-56b+73
x:14=c rest r
5.x=?
x:7=12rest r
x:8=10rest r
QUOTE(laura77 @ Nov 3 2009, 07:37 PM)
te rog ajuta-ma la aceste ex
1.afla x daca
x :2006=5rest r
2.afla x daca
x :4=a rest b
x :1 0=b rest a
3. afla r daca
a=1 x 2x .....x18 +2006
a:125=c rest r
4.x=?
x=84a-56b+73
x:14=c rest r
5.x=?
x:7=12rest r
x:8=10rest r
1.afla x daca
x :2006=5rest r
2.afla x daca
x :4=a rest b
x :1 0=b rest a
3. afla r daca
a=1 x 2x .....x18 +2006
a:125=c rest r
4.x=?
x=84a-56b+73
x:14=c rest r
5.x=?
x:7=12rest r
x:8=10rest r
Se aplica teorema impartirii cu rest:
D=deimpartit
I=impartitor
C=cat
R=rest
Atunci D=I*C+R, unde C este diferit de 0 (>0) iar 0<=R<C
Sa luam primul exercitiu:
1.afla x daca
x :2006=5rest r
x=2006*5+r unde r trebuie sa fie mai mic decat 5 , adica poate lua valorile 0, 1, 2, 3 sau 4
x=10030+0=10030
x=10030+1=10031
x=10030+2=10032
......
x=10030+4=10034
si asa mai departe...
Succes!
QUOTE(Matei @ Nov 3 2009, 08:49 PM)
Se aplica teorema impartirii cu rest:
D=deimpartit
I=impartitor
C=cat
R=rest
Atunci D=I*C+R, unde C este diferit de 0 (>0) iar 0<=R<C
Sa luam primul exercitiu:
1.afla x daca
x :2006=5rest r
x=2006*5+r unde r trebuie sa fie mai mic decat 5 , adica poate lua valorile 0, 1, 2, 3 sau 4
x=10030+0=10030
x=10030+1=10031
x=10030+2=10032
......
x=10030+4=10034
si asa mai departe...
Succes!
D=deimpartit
I=impartitor
C=cat
R=rest
Atunci D=I*C+R, unde C este diferit de 0 (>0) iar 0<=R<C
Sa luam primul exercitiu:
1.afla x daca
x :2006=5rest r
x=2006*5+r unde r trebuie sa fie mai mic decat 5 , adica poate lua valorile 0, 1, 2, 3 sau 4
x=10030+0=10030
x=10030+1=10031
x=10030+2=10032
......
x=10030+4=10034
si asa mai departe...
Succes!
multumesc,la ex 3 cum il aflu pe a/
QUOTE(laura77 @ Nov 3 2009, 09:07 PM)
multumesc,la ex 3 cum il aflu pe a/
La ex. 3 trebuie aflat r nu a !
buna seara
te rog sa imi explici si mie cateva exercitii
1. Afla suma ultimelor 100 de cifre ale nr.
A= 1x2x3x...........x2008x2009+2009
2.Stiind ca
n! =1x2x3x.....xn,aflati ultima cifra a nr 2009!+9002!
3.Un nr de trei sau mai multe cifre diferite este numit ondulat daca printre cifrele lui nu exista trei cifre alaturate a,b,c care sa satisfaca vreunul dintre sirurile de inegalitati a<b<c sau a>b >c.
Care este diferenta dintre cel mai mare si cel mai mic nr ondulat de patru cifre?
4.fie sirul1,8,15,22......
-aflati termenul de pe locul 15
-calculati suma primilor 10 termeni
5. mihai isi propune sa citeasca o carte care are 120 pagini.Daca in fiecare zi Mihai ar citi cu zece pagini mai putin decat in ziua precedenta el ar tremina de citit cartea in 4 zile.cate pagini trebuie sa citeasca Mihai in prima zi?
multumesc
te rog sa imi explici si mie cateva exercitii
1. Afla suma ultimelor 100 de cifre ale nr.
A= 1x2x3x...........x2008x2009+2009
2.Stiind ca
n! =1x2x3x.....xn,aflati ultima cifra a nr 2009!+9002!
3.Un nr de trei sau mai multe cifre diferite este numit ondulat daca printre cifrele lui nu exista trei cifre alaturate a,b,c care sa satisfaca vreunul dintre sirurile de inegalitati a<b<c sau a>b >c.
Care este diferenta dintre cel mai mare si cel mai mic nr ondulat de patru cifre?
4.fie sirul1,8,15,22......
-aflati termenul de pe locul 15
-calculati suma primilor 10 termeni
5. mihai isi propune sa citeasca o carte care are 120 pagini.Daca in fiecare zi Mihai ar citi cu zece pagini mai putin decat in ziua precedenta el ar tremina de citit cartea in 4 zile.cate pagini trebuie sa citeasca Mihai in prima zi?
multumesc
QUOTE(laura77 @ Nov 15 2009, 05:46 PM)
buna seara
te rog sa imi explici si mie cateva exercitii
1. Afla suma ultimelor 100 de cifre ale nr.
A= 1x2x3x...........x2008x2009+2009
2.Stiind ca
n! =1x2x3x.....xn,aflati ultima cifra a nr 2009!+9002!
3.Un nr de trei sau mai multe cifre diferite este numit ondulat daca printre cifrele lui nu exista trei cifre alaturate a,b,c care sa satisfaca vreunul dintre sirurile de inegalitati a<b<c sau a>b >c.
Care este diferenta dintre cel mai mare si cel mai mic nr ondulat de patru cifre?
4.fie sirul1,8,15,22......
-aflati termenul de pe locul 15
-calculati suma primilor 10 termeni
5. mihai isi propune sa citeasca o carte care are 120 pagini.Daca in fiecare zi Mihai ar citi cu zece pagini mai putin decat in ziua precedenta el ar tremina de citit cartea in 4 zile.cate pagini trebuie sa citeasca Mihai in prima zi?
multumesc
te rog sa imi explici si mie cateva exercitii
1. Afla suma ultimelor 100 de cifre ale nr.
A= 1x2x3x...........x2008x2009+2009
2.Stiind ca
n! =1x2x3x.....xn,aflati ultima cifra a nr 2009!+9002!
3.Un nr de trei sau mai multe cifre diferite este numit ondulat daca printre cifrele lui nu exista trei cifre alaturate a,b,c care sa satisfaca vreunul dintre sirurile de inegalitati a<b<c sau a>b >c.
Care este diferenta dintre cel mai mare si cel mai mic nr ondulat de patru cifre?
4.fie sirul1,8,15,22......
-aflati termenul de pe locul 15
-calculati suma primilor 10 termeni
5. mihai isi propune sa citeasca o carte care are 120 pagini.Daca in fiecare zi Mihai ar citi cu zece pagini mai putin decat in ziua precedenta el ar tremina de citit cartea in 4 zile.cate pagini trebuie sa citeasca Mihai in prima zi?
multumesc
Crezi ca aceste probleme sunt pentru clasa a V-a? Te rog sa citesti si aici:
http://www.utilecopii.ro/forum/index.php?s...st&p=901685
QUOTE(Matei @ Nov 16 2009, 12:22 PM)
Crezi ca aceste probleme sunt pentru clasa a V-a? Te rog sa citesti si aici:
http://www.utilecopii.ro/forum/index.php?s...st&p=901685
http://www.utilecopii.ro/forum/index.php?s...st&p=901685
sunt ex de cls a-V -a,primele 3 date de profesoara in clasa iar ultim.doua din culegerea mate 2000+ 9/10, cls a V-a
QUOTE(laura77 @ Nov 16 2009, 05:55 PM)
sunt ex de cls a-V -a,primele 3 date de profesoara in clasa iar ultim.doua din culegerea mate 2000+ 9/10, cls a V-a
sper ca acest ex e la nivelul cl a V-A
1.demonstrati ca nr abc scris in baza 10 se divide cu 3 daca a+b+c se divide cu 3.
off greu cu matematica
multumesc
QUOTE(laura77 @ Nov 16 2009, 06:13 PM)
sper ca acest ex e la nivelul cl a V-A
1.demonstrati ca nr abc scris in baza 10 se divide cu 3 daca a+b+c se divide cu 3.
off greu cu matematica
multumesc
1.demonstrati ca nr abc scris in baza 10 se divide cu 3 daca a+b+c se divide cu 3.
off greu cu matematica
multumesc
iti raspund eu , ca idee , sper sa nu fie gresita si sa ma bata matei :-s
numarul de forma abc il scriem ca : 100a + 10b + c
asociem numerele pt a obtine ceva favorabil noua si scriem 100a +10b + c = 99a + 9b + (a+b+c)
99a se divide cu 3
9b se divide cu 3
a+b+c se divide cu 3 (avem dat din enuntul problemei)
din cele 3 => abc se divide cu 3
QUOTE(Claudia94 @ Nov 16 2009, 09:18 PM)
iti raspund eu , ca idee , sper sa nu fie gresita si sa ma bata matei :-s
numarul de forma abc il scriem ca : 100a + 10b + c
asociem numerele pt a obtine ceva favorabil noua si scriem 100a +10b + c = 99a + 9b + (a+b+c)
99a se divide cu 3
9b se divide cu 3
a+b+c se divide cu 3 (avem dat din enuntul problemei)
din cele 3 => abc se divide cu 3
numarul de forma abc il scriem ca : 100a + 10b + c
asociem numerele pt a obtine ceva favorabil noua si scriem 100a +10b + c = 99a + 9b + (a+b+c)
99a se divide cu 3
9b se divide cu 3
a+b+c se divide cu 3 (avem dat din enuntul problemei)
din cele 3 => abc se divide cu 3
multumesc
QUOTE(Claudia94 @ Nov 16 2009, 09:18 PM)
iti raspund eu , ca idee , sper sa nu fie gresita si sa ma bata matei :-s
numarul de forma abc il scriem ca : 100a + 10b + c
asociem numerele pt a obtine ceva favorabil noua si scriem 100a +10b + c = 99a + 9b + (a+b+c)
99a se divide cu 3
9b se divide cu 3
a+b+c se divide cu 3 (avem dat din enuntul problemei)
din cele 3 => abc se divide cu 3
numarul de forma abc il scriem ca : 100a + 10b + c
asociem numerele pt a obtine ceva favorabil noua si scriem 100a +10b + c = 99a + 9b + (a+b+c)
99a se divide cu 3
9b se divide cu 3
a+b+c se divide cu 3 (avem dat din enuntul problemei)
din cele 3 => abc se divide cu 3
O declar pe Claudia94 - asistenta mea!
matei , scuze de deranj. ce vrea sa spuna n E(apartine) N* - {1} ? ceea ce am subliniat nu am mai intalnit , e minus nu e \ 
si nu inteleg la ce se refera 
ms anticipat
si nu inteleg la ce se refera ms anticipat
QUOTE(Claudia94 @ Dec 9 2009, 09:45 PM)
matei , scuze de deranj. ce vrea sa spuna n E(apartine) N* - {1} ? ceea ce am subliniat nu am mai intalnit , e minus nu e \ si nu inteleg la ce se refera
ms anticipat
si nu inteleg la ce se refera ms anticipat
N*=multimea nr. naturale nenule
- si \ sunt echivalene (unu si acelasi lucru), deci:
n apartine N*-{1} inseamna ca n aprtine diferentei dintre multimile N* si {1}, adica n apartine mutimii numerelor naturale nenule si nu apartine multimii {1}; De fapt n este un nr. natural, dar care nu poate fi 0 sau 1...
QUOTE(Matei @ Dec 9 2009, 10:06 PM)
N*=multimea nr. naturale nenule
- si \ sunt echivalene (unu si acelasi lucru), deci:
n apartine N*-{1} inseamna ca n aprtine diferentei dintre multimile N* si {1}, adica n apartine mutimii numerelor naturale nenule si nu apartine multimii {1}; De fapt n este un nr. natural, dar care nu poate fi 0 sau 1...
- si \ sunt echivalene (unu si acelasi lucru), deci:
n apartine N*-{1} inseamna ca n aprtine diferentei dintre multimile N* si {1}, adica n apartine mutimii numerelor naturale nenule si nu apartine multimii {1}; De fapt n este un nr. natural, dar care nu poate fi 0 sau 1...
multumesc
buna matei,
am o problema de matematica pe care nu o stiu rezolva :
afla cel mai mare numar de 2 cifre care impartit la 17 da restul 2
P.S:suntt cls a-5-a Ajuta-ma terog
am o problema de matematica pe care nu o stiu rezolva :
afla cel mai mare numar de 2 cifre care impartit la 17 da restul 2
P.S:suntt cls a-5-a Ajuta-ma terog
CITAT(Matei @ Oct 4 2009, 11:26 AM)
Se dau:
x+2y=17
y-3z=4
a )
Adunam, membru cu membru ecuatiile date initial:
(x+2y)+(y-3z)=17+4
x+2y+y-3z=21
x+3y-3z=21
b )
2*(x+2y)+(y-3z)=2*17+4
2x+4y+y-3z=38
2x+5y-3z=38
c )
3*(x+2y)+2*(y-3z)=3*17+2*4
(3x+6y)+(2y-6z)=51+8=59
d) Incercati!?...
Am folosit * pentru semnul inmultirii
x+2y=17
y-3z=4
a )
Adunam, membru cu membru ecuatiile date initial:
(x+2y)+(y-3z)=17+4
x+2y+y-3z=21
x+3y-3z=21
b )
2*(x+2y)+(y-3z)=2*17+4
2x+4y+y-3z=38
2x+5y-3z=38
c )
3*(x+2y)+2*(y-3z)=3*17+2*4
(3x+6y)+(2y-6z)=51+8=59
d) Incercati!?...
Am folosit * pentru semnul inmultirii
Ok, am inteles cum se fac primele, adica a, b, dar nu inteleg cum face la c, pentru ca mie imi da 850, iar in carte scrie ca rezultatul este 425.
Daca x+2y=17 si y-3z=4, avem exercitiul (2x+4y)(x+4y-6z), din care eu inteleg ca prima prima parte este in felul urmator:((2x+4y) se traduce in 2(x+2y)=217=34. Partea a doua inteleg ca este urmatoarea:(x+4y-6z), in care eu deduc ca avem un (x+2y) si 2(y-3z) ea ce inseamna ca avem 17 si 8, ca adunate ne dau 25. Cum ajung la rezultatul de 425?, pentru ca mie imi da 850.
Daca x+2y=17 si y-3z=4, avem exercitiul (2x+4y)(x+4y-6z), din care eu inteleg ca prima prima parte este in felul urmator:((2x+4y) se traduce in 2(x+2y)=217=34. Partea a doua inteleg ca este urmatoarea:(x+4y-6z), in care eu deduc ca avem un (x+2y) si 2(y-3z) ea ce inseamna ca avem 17 si 8, ca adunate ne dau 25. Cum ajung la rezultatul de 425?, pentru ca mie imi da 850.
jianbin0708
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