Răspuns:
Explicație pas cu pas:
30 / ( 2 a - 1 ) ( b + 2 ) → fractie echiunitara => numaratorul = numitorul
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( 2 a - 1 ) ( b + 2 ) = 30 ; a, b ∈ N
30 = 1 x 30 = 30 x 1 3 x 10 = 10 x 3 = 2 x 15 = 15 x 2 =
= 6 x 5 = 5 x 6
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2 a - 1 = 30 => 2 a = 31 => a = 31 / 2 ∈N
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2 a - 1 = 1 ⇒ 2 a = 2 ⇒ a = 1 ∈ N
b + 2 = 30 ⇒ b = 30 - 2 ⇒ b = 28 ∈ N
⇔ ( a, b ) = ( 1; 28 )
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2 a - 1 = 3 ⇒ 2 a = 4 ⇒ a = 2 ∈ N
b + 2 = 10 ⇒ b = 10 - 2 ⇒ b = 8 ∈ N ⇔ ( a, b ) ⇔ ( 2; 8)
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2 a - 1 = 10 => 2 a = 11 => a = 11 / 2∉N
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2 a - 1 = 5 ⇒ 2 a = 6 ⇒ a = 6 : 2 ⇒ a = 3 ∈ N
b + 2 = 6 ⇒ b = 6 - 2 ⇒ b = 4 ∈ N
⇔ ( a; b ) = ( 3; 4)
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2 a - 1 ≠ 6; 2 a = 7; a = 3,5∉N
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2 a - 1 = 15 ⇒ 2 a = 15 + 1 ⇒ a = 16 : 2 ⇒ a = 8 ∈ N
b + 2 = 2 ⇒ b = 2 - 2 ⇒ b = 0 ∈ N
⇔( a; b ) = ( 8; 0 )
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2 a - 1 ≠ 2 ; a ∉ N
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( a; b) = ( 1; 28 ); ( 2; 8); ( 3; 4 ); ( 8; 0 ) ∈ N ⇒ 4 solutii
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Verific: 30 / [ ( 2 × 1 - 1 ) ( 28 + 2 )] = 30/ ( 1 x 30 ) = 30 / 30
30 / ( 3 x 10 ) = 30 / 30 => fractie echiunitara
30 / ( 5 x 6 ) = 30 / 30 => fractie echiunitara
30 / ( 15 x 2 ) = 30 / 30 √
[tex]\it \dfrac{30}{(2a-1)(b+2)}=1 \Rightarrow (2a-1)(b+2)=30\ \ \ \ (1)\\ \\ 2a-1=impar \stackrel{(1)}{\Longrightarrow} b+2=par\ \ \ \ \ (2)\\ \\ (2a-1)(b+1)=30 \stackrel{(2)}{=}\ 1\cdot30=3\cdot30=5\cdot6=15\cdot2\Rightarrow\\ \\ \\ \Rightarrow \begin{cases}\it 2a-1\in\{1,3,5,15\}|_{+1}\Rightarrow 2a\in\{2,4,6,16\}|_{:2}\Rightarrow a\in\{1,\ 2,\ 3,\ 8\}\\ \\ \it b+2\in\{30.10,6,2\}|_{-2} \Rightarrow b\in\{28,\ 8,\ 4,\ 0\}\end{cases}[/tex]
[tex]\it (a,b)\in\{(1,\ 28),(2,\ 8),(3,\ 4).(8,0)\}[/tex]
