a: \(\Leftrightarrow2n-2+3⋮n-1\)
\(\Leftrightarrow n-1\in\left\{1;-1;3;-3\right\}\)
hay \(n\in\left\{2;0;4;-2\right\}\)
b \(\Leftrightarrow12n-9⋮3n+1\)
\(\Leftrightarrow12n+4-13⋮3n+1\)
\(\Leftrightarrow3n+1\in\left\{1;-1;13;-13\right\}\)
hay \(n\in\left\{0;-\dfrac{2}{3};4;-\dfrac{14}{3}\right\}\)
c: \(\Leftrightarrow n\left(n-1\right)+1⋮n-1\)
\(\Leftrightarrow n-1\in\left\{1;-1\right\}\)
hay \(n\in\left\{2;0\right\}\)
Nhìn cứ giống toán lớp 8,9 hơn lớp 6 đúng hok
a, \(2\left(n-1\right)+3⋮n-1\Rightarrow3⋮n-1\Leftrightarrow n-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
| n-1 | 1 | -1 | 3 | -3 |
| n | 2 | 0 | 4 | -2 |
b, \(4n-3⋮3n+1\Leftrightarrow12n-9⋮3n+1\)
\(\Leftrightarrow4\left(3n+1\right)-13⋮3n+1\Rightarrow-13⋮3n+1\Rightarrow3n+1\inƯ\left(-13\right)=\left\{\pm1;\pm13\right\}\)
| 3n+1 | 1 | -1 | 13 | -13 |
| n | 0 | loại | 4 | loại |
c, \(n^2-n+1=n\left(n-1\right)+1⋮n-1\Rightarrow n-1\inƯ\left(1\right)=\left\{\pm1\right\}\)
| n-1 | 1 | -1 |
| n | 2 | 0 |
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