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a, \(3x-8⋮x-4\)
\(3\left(x-4\right)+4⋮x-4\)
\(4⋮x-4\)hay \(x-4\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
| x - 4 | 1 | -1 | 2 | -2 | 4 | -4 |
| x | 5 | 3 | 6 | 2 | 8 | 0 |
c, tương tự
a,Gợi ý:vì x^2+x+1 chia hết cho x+1 => x^2 chia hết cho x+1 b,Gợi ý nhân 3 với (x-4) rồi lấy 3x-8 trừ đi c,lấy (x+5) trừ đi x-2 e,Gợi ý x^2+2x-7 chia hết cho x+2
a, x+3 chia hết cho x-1
Ta có: x+3=(x+1)+2
=> 2 chia hết cho x+1
=>x+1 thuộc Ư(2)= {1, -1, 2, -2}
=> x thuộc {0,-2, 1, -3}
b.
b,3x chia hết cho x-1
c,2-x chia hết cho x+1
Ta có:
\(\dfrac{x+3}{x-1}=\dfrac{x-1+4}{x-1}=1+\dfrac{4}{x-1}\)
Để (x + 3) \(⋮\left(x-1\right)\) thì 4 \(⋮\left(x-1\right)\)
\(\Rightarrow\) x - 1 = 1; x - 1 = -1; x - 1 = 2; x - 1 = -2; x - 1 = 4; x - 1 = -4
*) x - 1 = 1
x = 2
*) x - 1 = -1
x = 0
*) x - 1 = 2
x = 3
*) x - 1 = -2
x = -1
*) x - 1 = 4
x = 5
*) x - 1 = -4
x = -3
Vậy x = 5; x = 3; x = 2; x = 0; x = -1; x = -3
b: \(x^2+2x-7⋮x+2\)
\(\Leftrightarrow x\left(x+2\right)-7⋮x+2\)
\(\Leftrightarrow x+2\in\left\{1;-1;7;-7\right\}\)
hay \(x\in\left\{-1;-3;5;-9\right\}\)
c: \(x^2+x+1⋮x+1\)
\(\Leftrightarrow x\left(x+1\right)+1⋮x+1\)
\(\Leftrightarrow x+1\in\left\{1;-1\right\}\)
hay \(x\in\left\{0;-2\right\}\)
e: \(x+5⋮x-2\)
\(\Leftrightarrow x-2+7⋮x-2\)
\(\Leftrightarrow x-2\in\left\{1;-1;7;-7\right\}\)
hay \(x\in\left\{3;1;9;-5\right\}\)
a)<=>(x+1)+2 chia hết x+1
=>2 chia hết x+1
=>x+1\(\in\){1,-1,2,-2}
=>x\(\in\){0,-2,1,-3}
b)<=>3(x-2)+7 chia hết x-2
=>7 chia hết x-2
=>x-2\(\in\){1,-1,7,-7}
=>x\(\in\){3,1,9,-5}
c,d,e tương tự
a, x+8 chia hết cho x+7
=>x+7+1 chia hết cho x+7
=>1 chia hết cho x+7
=> x+7=1hoặc -1
=>x=(-6) hoặc (-8)
b, 2x+16 chia hết cho x+7
2(x+7)+2 chia hết cho x+7
.....
c,mọi số x
d,6 ,4
d,2,0,-2,-4
click dúng nhớ
a)Ta có : \(x-5⋮x+2=>x-5-\left(x+2\right)⋮x-2=>-7⋮x-2\)
\(=>x-2\inƯ\left(7\right)\left\{-7;-1;1;7\right\}\)
\(=>x\in\left\{-5;1;3;9\right\}\)
b)Ta có : \(2x+1⋮2x-1=>2x+1-\left(2x-1\right)⋮2x-1=>2⋮2x-1\)
\(=>2x-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
\(=>2x\in\left\{-1;0;2;3\right\}\)
\(=>x\in\left\{0;1\right\}\)(vì \(x\in Z\))
c)\(\left(x+5\right)-3\left(x+5\right)+2⋮x+5=>2⋮x+5=>x+5\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
\(=>x\in\left\{-7;-6;-4;-3\right\}\)
d)\(x+1⋮x+2=>x+2-1⋮x+2\)
\(=>1⋮x+2=>x+2\inƯ\left(1\right)=\left\{1;-1\right\}=>x\in\left\{-1;-3\right\}\)
a) \(Ư\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
Suy ra \(x\in\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
b) \(Ư\left(13\right)=\left\{\pm1;\pm13\right\}\)
| x + 1 | 1 | 13 | -1 | -13 |
| x | 0 | 12 | -2 | -14 |
Suy ra \(x\in\left\{0;12;-2;-14\right\}\)
c) Số nào chia hết cho x - 3 vậy????
d) \(\left(x+8\right)⋮\left(x+2\right)\Leftrightarrow\left(x+2+6\right)⋮\left(x+2\right)\)
Mà x + 2 chia hết cho x + 2 nên 6 chia hết cho x + 2
\(Ư\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
| x + 2 | 1 | 2 | 3 | 6 | -1 | -2 | -3 | -6 |
| x | -1 | 0 | 1 | 4 | -3 | -4 | -5 | -8 |
Suy ra \(x\in\left\{-1;0;1;4;-3;-4;-5;-8\right\}\)
5.
$4x+3\vdots x-2$
$\Rightarrow 4(x-2)+11\vdots x-2$
$\Rightarrow 11\vdots x-2$
$\Rightarrow x-2\in \left\{1; -1; 11; -11\right\}$
$\Rightarrow x\in \left\{3; 1; 13; -9\right\}$
6.
$3x+9\vdots x+2$
$\Rightarrow 3(x+2)+3\vdots x+2$
$\Rightarrow 3\vdots x+2$
$\Rightarrow x+2\in \left\{1; -1; 3; -3\right\}$
$\Rightarrow x\in \left\{-1; -3; 1; -5\right\}$
7.
$3x+16\vdots x+1$
$\Rightarrow 3(x+1)+13\vdots x+1$
$\Rightarrow 13\vdots x+1$
$\Rightarrow x+1\in \left\{1; -1; 13; -13\right\}$
$\Rightarrow x\in\left\{0; -2; 12; -14\right\}$
8.
$4x+69\vdots x+5$
$\Rightarrow 4(x+5)+49\vdots x+5$
$\Rightarrow 49\vdots x+5$
$\Rightarrow x+5\in\left\{1; -1; 7; -7; 49; -49\right\}$
$\Rightarrow x\in \left\{-4; -6; 2; -12; 44; -54\right\}$
** Bổ sung điều kiện $x$ là số nguyên.
1. $x+9\vdots x+7$
$\Rightarrow (x+7)+2\vdots x+7$
$\Rightarrow 2\vdots x+7$
$\Rightarrow x+7\in \left\{1; -1; 2; -2\right\}$
$\Rightarrow x\in \left\{-6; -8; -5; -9\right\}$
2. Làm tương tự câu 1
$\Rightarrow 9\vdots x+1$
3. Làm tương tự câu 1
$\Rightarrow 17\vdots x+2$
4. Làm tương tự câu 1
$\Rightarrow 18\vdots x+2$