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b) \(3x+9=3x+6+3=3\left(x+2\right)+3⋮\left(x+2\right)\Leftrightarrow3⋮\left(x+2\right)\)
\(\Leftrightarrow x+2\inƯ\left(3\right)=\left\{-3,-1,1,3\right\}\Leftrightarrow x\in\left\{-5,-3,-1,1\right\}\).
a), c) tương tự.
d) \(\left(2x+1\right)⋮\left(3x-1\right)\Rightarrow3\left(2x+1\right)=6x+3=6x-2+5=2\left(3x-1\right)+5⋮\left(3x-1\right)\)
\(\Leftrightarrow5⋮\left(3x-1\right)\Leftrightarrow3x-1\inƯ\left(5\right)=\left\{-5,-1,1,5\right\}\Leftrightarrow x\in\left\{0,2\right\}\)(vì \(x\)nguyên)
Thử lại đều thỏa mãn.
a) Ta có: \(2x-2\)\(⋮\)\(x-2\)
\(\Leftrightarrow\)\(2\left(x-2\right)+2\)\(⋮\)\(x-2\)
Ta thấy \(2\left(x-2\right)\)\(⋮\)\(x-2\)
nên \(2\)\(⋮\)\(x-2\)
hay \(x-2\)\(\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Ta lập bảng sau:
\(x-2\) \(-2\) \(-1\) \(1\) \(2\)
\(x\) \(0\) \(1\) \(3\) \(4\)
Vậy \(x=\left\{0;1;3;4\right\}\)
a) Để \(38-3x⋮x\)mà \(3x⋮x\)
\(\Rightarrow\)\(38⋮x\)\(\Rightarrow\)\(x\inƯ\left(38\right)\in\left\{\pm1;\pm2;\pm9;\pm38\right\}\)
Vì \(x\inℕ\)\(\Rightarrow\)\(x\in\left\{1;2;9;38\right\}\)
Vậy \(x\in\left\{1;2;9;38\right\}\)
b) Ta có: \(3x+7=\left(3x-3\right)+10=3.\left(x-1\right)+10\)
- Để \(3x+7⋮x-1\)\(\Leftrightarrow\)\(3.\left(x-1\right)+10⋮x-1\)mà \(3.\left(x-1\right)⋮x-1\)
\(\Rightarrow\)\(10⋮x-1\)\(\Rightarrow\)\(x-1\inƯ\left(10\right)\in\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
- Ta có bảng giá trị:
| \(x-1\) | \(-1\) | \(1\) | \(-2\) | \(2\) | \(-5\) | \(5\) | \(-10\) | \(10\) |
| \(x\) | \(0\) | \(2\) | \(-1\) | \(3\) | \(-4\) | \(6\) | \(-9\) | \(11\) |
| \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(L\right)\) | \(\left(TM\right)\) | \(\left(L\right)\) | \(\left(TM\right)\) | \(\left(L\right)\) | \(\left(TM\right)\) |
( Loại vì \(x\inℕ\))
Vậy \(x\in\left\{0;2;3;6;11\right\}\)
c) Ta có: \(2x+19=\left(2x+1\right)+18\)
- Để \(2x+19⋮2x+1\)\(\Leftrightarrow\)\(\left(2x+1\right)+18⋮2x+1\)mà \(2x+1⋮2x+1\)
\(\Rightarrow\)\(18⋮2x+1\)\(\Rightarrow\)\(2x+1\inƯ\left(18\right)\in\left\{\pm1;\pm2;\pm3;\pm6;\pm9;\pm18\right\}\)
Vì \(2x+1\)là lẻ \(\Rightarrow\)\(2x+1\in\left\{\pm1;\pm3;\pm9\right\}\)
- Ta có bảng giá trị:
| \(2x+1\) | \(-1\) | \(1\) | \(-3\) | \(3\) | \(-9\) | \(9\) |
| \(x\) | \(-1\) | \(0\) | \(-2\) | \(1\) | \(-5\) | \(4\) |
| \(\left(L\right)\) | \(\left(TM\right)\) | \(\left(L\right)\) | \(\left(TM\right)\) | \(\left(L\right)\) | \(\left(TM\right)\) |
( loại vì \(x\inℕ\))
Vậy \(x\in\left\{0;1;4\right\}\)
a)\(\frac{x+11}{x-6}=\frac{x-6+17}{x-6}=\frac{x-6}{x-6}+\frac{17}{x-6}\)
=>x-6\(\in\) Ư(17)
| x-6 | 1 | -1 | 17 | -17 |
| x | 7 | 5 | 23 | -11 |