Toán Tìm a E Z biết $a^{2}$ + 5 chia hết cho a + 1 13/07/2021 By Genesis Tìm a E Z biết $a^{2}$ + 5 chia hết cho a + 1
Có $a^{2} + 5 = a^{2} – a + a + 5 = a(a – 1) + (a – 1) + 6 = (a – 1)(a + 1) + 6$ $\Rightarrow a^{2} + 5 \vdots a + 1$ $\Leftrightarrow (a – 1)(a + 1) + 6 \vdots a + 1$ $\Leftrightarrow 6 \vdots a + 1$ $\Leftrightarrow a + 1 \in Ư(6) = \left \{ -6; -3; -2; -1; 1; 2; 3; 6 \right \}$ $\Rightarrow a \in \left \{ -7; -4; -3; -2; 0; 1; 2; 5 \right \}$ Trả lời
Ta có: a²+5$\vdots$a+1 ⇒(a²-1)+6$\vdots$a+1 ⇒(a-1)(a+1)+6$\vdots$a+1 ⇒a+1∈Ư(6)={±1;±2;±3;±6} a+1 1 -1 2 -2 3 -3 6 -6 a 0 -2 1 -3 2 -4 5 -7 Vậy a∈{0;-2;1;-3;2;-4;5;-7} Trả lời
Có $a^{2} + 5 = a^{2} – a + a + 5 = a(a – 1) + (a – 1) + 6 = (a – 1)(a + 1) + 6$
$\Rightarrow a^{2} + 5 \vdots a + 1$
$\Leftrightarrow (a – 1)(a + 1) + 6 \vdots a + 1$
$\Leftrightarrow 6 \vdots a + 1$
$\Leftrightarrow a + 1 \in Ư(6) = \left \{ -6; -3; -2; -1; 1; 2; 3; 6 \right \}$
$\Rightarrow a \in \left \{ -7; -4; -3; -2; 0; 1; 2; 5 \right \}$
Ta có: a²+5$\vdots$a+1
⇒(a²-1)+6$\vdots$a+1
⇒(a-1)(a+1)+6$\vdots$a+1
⇒a+1∈Ư(6)={±1;±2;±3;±6}
a+1 1 -1 2 -2 3 -3 6 -6
a 0 -2 1 -3 2 -4 5 -7
Vậy a∈{0;-2;1;-3;2;-4;5;-7}