Question

Tìm n:

4n - 5 chia hết cho 2n - 1

Help Me

Answer 1

Ta có:

\(\frac{4n-5}{2n-1}=\frac{2\left(2n-1\right)-3}{2n-1}=2-\frac{3}{2n-1}\)

Để (4n - 5) \(⋮\) (2n - 1) thì 3 \(⋮\) (2n - 1)

\(\Rightarrow\) 2n - 1 = -1; 2n - 1 = 1; 2n - 1 = 3; 2n - 1 = -3

*) 2n - 1 = 1

2n = 2

n = 1

*) 2n - 1 = -1

2n = 0

n = 0

*) 2n - 1 = 3

2n = 4

n = 2

*) 2n - 1 = -3

2n = -2

n = -1

Vậy n = -1; n = 2; n = 0; n = 1

Answer 2

4n-5⋮2n-1

4n-2-3⋮2n-1

2.2n-2.1-3⋮2n-1

2(2n-1)-3⋮2n-1

Vì 2(2n-1)-3⋮2n-1

2n-1⋮2n-1 => 2(2n-1)⋮2n-1

=>3⋮2n-1

=>2n-1∈ Ư(3)

=>2n-1=1; -1; 3; -3

Ta có bảng sau:

2n-1	1	-1	3	-3
n	1	0	2	-1

Vậy n=1; 0; 2; -1.