Câu `C` là `(4x+4)/(2x)` cộng với bao nhiêu ạ?
`A = (4x+9)/(2x+1)`
Đk: `x ne -1/2`
`(4x+9)/(2x+1) = (4x+2)/(2x+1) + 7/(2x+1) = 2 + 7/(2x+1) in ZZ`.
`-> 7 vdots 2x + 1 -> 2x + 1 in Ư(7)`
`-> 2x + 1 in {+-1, +-7}`
`-> x in {0, -1, 3, -4}`
`C = (4x+4)/(2x+4)`
`= (2x+2)/(x+2)`
`= (2x+4)/(x+2) - 2/(x+2) in ZZ`
`-> 2 vdots (x+2)`
`-> x + 2 in Ư(2)`
`-> x + 2 in {+-1, +-2}`
`-> x in {-3, -1, 0, -4}`.
\(A=\dfrac{4x+9}{2x+1}=\dfrac{2\left(2x+1\right)+7}{2x+1}=2+\dfrac{7}{2x+1}\)
\(Để A\in Z\) \(=>\dfrac{7}{2x+1}\in Z\)
\(=>7⋮\left(2x+1\right)=>2x+1\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
| \(2x+1\) | \(-1\) | \(1\) | \(-7\) | \(7\) |
| \(2x\) | \(-2\) | \(0\) | \(-8\) | \(6\) |
| \(x\) | \(-1\left(TM\right)\) | \(0\left(TM\right)\) | \(-4\left(TM\right)\) | \(3\left(TM\right)\) |
\(ĐK:x\ne-2\)
\(C=\dfrac{4x+4}{2x+4}=\dfrac{2\left(2x+4\right)-4}{2x+4}=2-\dfrac{4}{2x+4}\)
\(Để\ C\in Z\) \(=>\dfrac{4}{2x+4}\in Z=>4⋮\left(2x+4\right)=>2x+4\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
| \(2x+4\) | \(-1\) | \(1\) | \(-2\) | \(2\) | \(-4\) | \(4\) |
| \(2x\) | \(-5\) | \(-3\) | \(-6\) | \(-2\) | \(-8\) | \(0\) |
| \(x\) | \(-\dfrac{5}{2}\left(KTM\right)\) | \(-\dfrac{3}{2}\left(KTM\right)\) | \(-3\left(TM\right)\) | \(-1\left(TM\right)\) | \(-4\left(TM\right)\) | \(0\left(TM\right)\) |