Để \(\frac{2x+5}{x+1}\)là số tự nhiên
\(\Rightarrow2x+5⋮x+1\)
\(\Rightarrow2x+2+3⋮x+1\)
\(\Rightarrow2\left(x+1\right)+3⋮x+1\)
mà \(2\left(x+1\right)⋮x+1\Rightarrow3⋮x+1\)
\(\Rightarrow x+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Nếu : x + 1 = 1 => x = 0 ( TM )
x + 1 = -1 => x = -2 ( loại )
x + 1 = 3 => x = 2 ( TM )
x + 1 = -3 => x = -4 ( loại )
\(\Rightarrow x\in\left\{0;2\right\}\)
\(a,\left(2x+1\right)\left(3y-2\right)=12\)
\(\Rightarrow2x+1;3x-2\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
.... như bài 1
\(b,1+2+3+..+x=55.\)
\(\Rightarrow\left(x+1\right).x:2=55\)
\(\Rightarrow\left(x+1\right).x=110\)
mà \(x\left(x+1\right)=10.11\)
\(\Rightarrow x=10\)
1/ để \(\frac{2x+5}{x+1}\)là số tự nhiên thì
\(2x+5⋮x+1\)
\(\Rightarrow2x+2+3⋮x+1\)
\(\Rightarrow2\left(x+1\right)+3⋮x+1\)
vì \(2\left(x+1\right)⋮x+1\)
\(\Rightarrow3⋮x+1\)
\(\Rightarrow x+1\inƯ\left(3\right)=\left\{1;3\right\}\)
\(\Rightarrow x\in\left\{0;2\right\}\)
2/ a) \(\left(2x+1\right)\left(3y-2\right)=12\)
\(\Rightarrow\left(2x+1\right);\left(3y-2\right)\inƯ\left(12\right)\)
Xét bảng sau:
| 2x + 1 | 1 | 12 | 2 | 6 | 3 | 4 |
| 3y - 2 | 12 | 1 | 6 | 2 | 4 | 3 |
| x | 0 | \(\frac{11}{2}\) | \(\frac{1}{2}\) | \(\frac{5}{2}\) | 1 | \(\frac{3}{2}\) |
| y | \(\frac{14}{3}\) | 1 | \(\frac{8}{3}\) | \(\frac{4}{3}\) | 2 | \(\frac{5}{3}\) |
vậy các cặp số x;y lần lượt là (0 ; 14/3) ; (11/2 ; 1) ; (1/2 ; 8/3) ; (5/2 ; 4/3) ; (1 ; 2) ; (3/2 ; 5/3)
