1. |x+1| < 2 - 3 ≤ |x- 2| ≤ 3 2 ≤ |x- 5 | ≤ 4 8 < |x - 6 | ≤ 11 2. 15 ⋮ ( x + 7 )

Giải thích các bước giải:

1.aTa có:

$|x+1|<2\to -2<x+1<2\to -3<x<1$

b.Ta có:

$-3\le |x-2|\le 3$

Vì $|x-2|\ge 0,\quad\forall x$

$\to |x-2|\ge -3$ luôn đúng

$\to |x-2|\le 3$

$\to -3\le x-2\le 3$

$\to -1\le x\le 5$

c.Ta có:

$2\le |x-5|\le 4$

$+)2\le |x-5|$

$\to x-5\ge 2\to x\ge 7$ hoặc $x-5\le -2\to x\le 3$

$+)|x-5|\le 4\to -4\le x-5\le 4\to 1\le x\le 9$

Kết hợp cả $2$ trường hợp

$\to 2\le abs(x-5)\le 4$

$\to 1\le x\le 3$

Hoặc $7\le x\le 9$

d.Ta có:

$8<|x-6|\le 11$

$+)8<|x-6|$

$\to 8<x-6\to x>14$

Hoặc $x-6<-8\to x<-2$

$+)|x-6|\le 11\to -11\le x-6\le 11\to -5\le x\le 17$

Kết hợp cả $2$ trường hợp
$\to -5\le x<-2$ và $14<x\le 17$

Bài 2:

a.Ta có:

$15\quad\vdots\quad x+7$

$\to x+7\in U(15)$

$\to x+7\in\{1,3,5,15,-1,-3,-5,-15\}$

$\to x\in\{-6,-4,-2,8,-8,-10,-12,-22\}$

b.Ta có:

$3x+2\quad\vdots\quad x-1$

$\to 3x-3+3+2\quad\vdots\quad x-1$

$\to 3(x-1)+5\quad\vdots\quad x-1$

$\to 5\quad\vdots\quad x-1$

$\to x-1\in U(5)$

$\to x-1\in\{1,5,-1,-5\}$

$\to x\in\{2,6,0,-4\}$

c.Ta có:

$x+5\quad\vdots\quad x-2$

$\to x-2+2+5\quad\vdots\quad x-2$

$\to x-2+7\quad\vdots\quad x-2$

$\to 7\quad\vdots\quad x-2$

$\to x-2\in\{1,7,-1,-7\}$

$\to x\in\{3,9,1,-5\}$

d.Ta có:
$x^2+x+2\quad\vdots\quad x+1$

$\to (x^2+x)+2\quad\vdots\quad x+1$

$\to x(x+1)+2\quad\vdots\quad x+1$

$\to 2\quad\vdots\quad x+1$

$\to x+1\in\{1,2,-1,-2\}$

$\to x\in\{0,1,-2,-3\}$

3.a.Ta có:

$|x+1|\ge 0, |x-2|\ge 0, |x+7|\ge 0,\quad\forall x$

$\to |x+1|+|x-2|+|x+7|\ge 0$

$\to5x-10\ge 0$

$\to 5x\ge 10$

$\to x\ge 2$

b.Ta có $x\ge 2$

$\to x+1\ge 2+1>0\to |x+1|=x+1$

      $x-2\ge 2-2=0\to |x-2|=x-2$

      $x+7\ge 2+7>0\to |x+7|=x+7$

Khi đó phương trình đã cho trở thành:

$x+1+x-2+x+7=5x-10$

$\to 3x+6=5x-10$

$\to 2x=16$

$\to x=8$

Trả lời