Question

Tìm số nguyên x để các phấ số sau có giá trị là 1 sô nguyên

A  =4x + 9/2x + 1

C = 4x + 4/2x + 

Answer 1

Câu `C` là `(4x+4)/(2x)` cộng với bao nhiêu ạ?

Answer 2

`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}`.

Answer 3

\(A=\dfrac{4x+9}{2x+1}=\dfrac{2\left(2x+1\right)+7}{2x+1}=2+\dfrac{7}{2x+1}\)

\(Để A\in Z\) \(=>\dfrac{7}{2x+1}\in Z\)

\(=>7⋮\left(2x+1\right)=>2x+1\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(2x+1\)	\(-1\)	\(1\)	\(-7\)	\(7\)
\(2x\)	\(-2\)	\(0\)	\(-8\)	\(6\)
\(x\)	\(-1\left(TM\right)\)	\(0\left(TM\right)\)	\(-4\left(TM\right)\)	\(3\left(TM\right)\)

 

 

 

Answer 4

\(ĐK:x\ne-2\)

\(C=\dfrac{4x+4}{2x+4}=\dfrac{2\left(2x+4\right)-4}{2x+4}=2-\dfrac{4}{2x+4}\)

\(Để\ C\in Z\) \(=>\dfrac{4}{2x+4}\in Z=>4⋮\left(2x+4\right)=>2x+4\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

\(2x+4\)	\(-1\)	\(1\)	\(-2\)	\(2\)	\(-4\)	\(4\)
\(2x\)	\(-5\)	\(-3\)	\(-6\)	\(-2\)	\(-8\)	\(0\)
\(x\)	\(-\dfrac{5}{2}\left(KTM\right)\)	\(-\dfrac{3}{2}\left(KTM\right)\)	\(-3\left(TM\right)\)	\(-1\left(TM\right)\)	\(-4\left(TM\right)\)	\(0\left(TM\right)\)