a) Để \(-5:\left(x-4\right)\)là số nguyên
\(\Rightarrow x-4\inƯ\left(-5\right)\in\left\{\pm1; \pm5\right\}\)
- Ta có bảng giá trị:
\(x-4\) | \(-1\) | \(1\) | \(-5\) | \(5\) |
\(x\) | \(3\) | \(5\) | \(-1\) | \(9\) |
\(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) |
Vậy \(x\in\left\{-1; 3; 5; 9\right\}\)
b) Ta có: \(x+8=\left(x+7\right)+1\)
- Để \(x+8⋮x+7\)\(\Rightarrow\)\(\left(x+7\right)+1⋮x+7\)mà \(x+7⋮x+7\)
\(\Rightarrow\)\(1⋮x+7\)\(\Rightarrow\)\(x+7\inƯ\left(1\right)\in\left\{\pm1\right\}\)
+ \(x+7=1\)\(\Leftrightarrow\)\(x=1-7=-6\left(TM\right)\)
+ \(x+7=-1\)\(\Leftrightarrow\)\(x=-1-7=-8\left(TM\right)\)
Vậy \(x\in\left\{-1; -8\right\}\)
c) Ta có: \(2x-9=\left(2x-10\right)+1=2.\left(x-5\right)+1\)
- Để \(2x-9⋮x-5\)\(\Rightarrow\)\(2.\left(x-5\right)+1⋮x-5\)mà \(2.\left(x-5\right)⋮ x-5\)
\(\Rightarrow\)\(1⋮x-5\)\(\Rightarrow\)\(x-5\inƯ\left(1\right)\in\left\{\pm1\right\}\)
+ \(x-5=1\)\(\Leftrightarrow\)\(x=1+5=6\left(TM\right)\)
+ \(x-5=-1\)\(\Leftrightarrow\)\(x=-1+5=4\left(TM\right)\)
Vậy \(x\in\left\{4; 6\right\}\)
d) Ta có: \(5x+2=\left(5x+5\right)-3=5.\left(x+1\right)-3\)
- Để \(5x+2⋮x+1\)\(\Rightarrow\)\(5.\left(x+1\right)-3⋮x+1\)mà \(5.\left(x+1\right)⋮x+1\)
\(\Rightarrow\)\(3⋮x+1\)\(\Rightarrow\)\(x+1\inƯ\left(3\right)\in\left\{\pm1; \pm3\right\}\)
- Ta có bảng giá trị:
\(x+1\) | \(-1\) | \(1\) | \(-3\) | \(3\) |
\(x\) | \(-2\) | \(0\) | \(-4\) | \(2\) |
\(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) |
Vậy \(x\in\left\{-4;-2; 0; 2\right\}\)