1/ ( x^2 -3x -5 ) chia hết cho (x-3) 2/(5x+2) chia hết cho(x+1) 3/(2x^2 + 3x+2) chia hết cho (x+1) 16/10/2021 Bởi Mackenzie 1/ ( x^2 -3x -5 ) chia hết cho (x-3) 2/(5x+2) chia hết cho(x+1) 3/(2x^2 + 3x+2) chia hết cho (x+1)
1/ $x²-3x-5\vdots x-3$ $→x(x-3)-5\vdots x-3$ Do $x(x-3)\vdots x-3$ $→-5\vdots x-3$ $→x-3\in Ư(-5)=\{±1;±5\}$ $→x\in \{4;2;8;-2\}$ 2/ $5x+2\vdots x+1$ $↔5(x+1)-3\vdots x+1$ Do $5(x+1)\vdots x+1$ $→3\vdots x+1$ $→x+1\in Ư(3)=\{±1;±3\}$ $→x\in \{0;-2;2;-4\}$ 3/ $2x²+3x+2\vdots x+1$ $→2x(x+1)+x+1+1\vdots x+1$ $→(x+1)(2x+1)+1\vdots x+1$Do $(x+1)(2x+1)\vdots x+1$ $→1\vdots x+1$ $→x+1\in Ư(1)=\{±1\}$ $→x\in\{0;-2\}$ Bình luận
1/ $x²-3x-5\vdots x-3$
$→x(x-3)-5\vdots x-3$
Do $x(x-3)\vdots x-3$
$→-5\vdots x-3$
$→x-3\in Ư(-5)=\{±1;±5\}$
$→x\in \{4;2;8;-2\}$
2/ $5x+2\vdots x+1$
$↔5(x+1)-3\vdots x+1$
Do $5(x+1)\vdots x+1$
$→3\vdots x+1$
$→x+1\in Ư(3)=\{±1;±3\}$
$→x\in \{0;-2;2;-4\}$
3/ $2x²+3x+2\vdots x+1$
$→2x(x+1)+x+1+1\vdots x+1$
$→(x+1)(2x+1)+1\vdots x+1$
Do $(x+1)(2x+1)\vdots x+1$
$→1\vdots x+1$
$→x+1\in Ư(1)=\{±1\}$
$→x\in\{0;-2\}$