Question

Determinati multimile:
A={x∈Z/6x+4=4x+1};
B={x∈N/x² +2x ≤ 0}
C={x/3x-1≤ 2x+4, x∈N};
D={(x,y)/x∈N , y∈ N si 3x+2y=12};
$E=\{\overline{ab}/7*\overline{ab}=4*\overline{ba}\}.$

Answer 1

A 
6x  -4x = 1 -4   ;    2x =  -3   ; x= -3 /2 ∉ Z         A=Ф multimea vida 
B  x( x+ 2) ≤ 0  
x( ( x+2) < 0 fals , numerele naturale inmultite nu sunt mai mici ca 0 
dar x( x+2) = 0  daca x=0 ∈N      B= {0}
C    3x -2x ≤ 4 +1 ;        x ≤ 5  naturale  C= { 0,1,2,3,4,5 }
D        x∈N    =       0         1           2          3          4  (am dat valori )
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         y∈N      =       6         fals       3          fals       0
D= { (0,6) ; (2,3);(4,0)
E  a , b =1,2,3,4,5,6,7,8,9, cifrele unui numar 
ab se imparte la 7 
ab= 14,21,28,35,42,49,56,63,77,84,91,98 
14          2· 14 =4· 41 fals 
21           3·21=4·12    fals 
28            4·28 =4·82  fals
35            5·35 =4·53 fals 
42            6·42 = 4· 24 fals 
49            7·49 = 4· 94fals 
56            8·56 = 4· 65 fals 
63            9·63=4·36 fals 
77           11·77 = 4·77 fals

Answer 2

$6x+4=4x+1\ \textless \ =\ \textgreater \ 2x=-3=\ \textgreater \ x=- \frac{3}{2}$∉Z.

A=Ф (multimea vida)
*De altfel, daca x era intreg, atunci membrul stang era par, iar cel drept era impar.*
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$x^{2} +2x \leq 0\ \textless \ =\ \textgreater \ x^{2} +2x+1 \leq 1\ \textless \ =\ \textgreater \ (x+1) ^{2} \leq 1=\ \textgreater$ (x+1)∈[-1;1] => x∈[-2;0], dar x∈N => x=0.

B={0}.
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$3x-1 \leq 2x+4\ \textless \ =\ \textgreater \ x \leq 5.$ Cum x∈N=> x∈{0;1;2;3;4;5} => C={0;1;2;3;4;5}.
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3x+2y=12=>3x=12-2y<=> 3x=2(6-y). Dar 2 si 3 sunt prime intre ele => 3 | 6-y si 2 | x. Mai mult, 3 | 6-y implica 3 | y.

Deci exista a,b∈N astfel incat x=2a si y=3b.
3x+2y=12<=>6a+6b=12 <=> a+b=2 => (a,b)∈{(0;2);(1;1);(2;0) => (x;y)∈{(0;6);(2;3);(4;0)}.

D={(0;6);(2;3);(4;0)}.
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$7*\overline{ab}=4*\overline{ba}\ \textless \ =\ \textgreater \ 7(10a+b)=4(10b+a)\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ 70a+7b=40b+4a\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ 66a=33b\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ 2a=b.$

Deci $\overline{ab}$∈{21; 42; 63; 84}.

E={21; 42; 63; 84}.