Question

[Kudo Shinichi]

Tìm các số tự nhiên x biết :

a) x + 20 là B( x + 2 )

b) x + 5 là Ư(4x + 69 )

c) 10x + 23 chia hết cho 2x + 1

Answer 1

[Nguyễn Nam]

c) Ta có:

\(10x+23⋮2x+1\)

\(\Rightarrow\left(10x+5\right)+18⋮2x+1\)

\(\Rightarrow5\left(2x+1\right)+18⋮2x+1\)

\(\Rightarrow18⋮2x+1\)

\(\Rightarrow2x+1\inƯ\left(18\right)=\left\{1;2;3;6;9;18\right\}\) ( vì \(x\in N\) )

+) \(2x+1=1\Rightarrow x=0\left(thoa\right)\)

+) \(2x+1=2\Rightarrow x=0,5\left(loai\right)\)

+) \(2x+1=3\Rightarrow x=1\left(thoa\right)\)

+) \(2x+1=6\Rightarrow x=2,5\left(loai\right)\)

+) \(2x+1=1\Rightarrow x=4\left(thoa\right)\)

+) \(2x+1=1\Rightarrow x=8,5\left(loai\right)\)

Vậy \(x\in\left\{0;1;4\right\}\)

Answer 2

[Nguyen Thi Huyen]

a) Ta có: x + 20 = (x + 2) + 18 => (x +2) + 18 \(⋮\) (x + 2) khi 18 \(⋮\) (x + 2)

=> x + 2 \(\in\) Ư(18) = {1; 2; 3; 6; 9; 18}

Vì x \(\in\) N

=> x \(\in\) {0; 1; 4; 7; 16}

b) Ta có: x + 5 \(⋮\) (x + 5)

=> 4x + 20 \(⋮\) (x + 5)

Và 4x + 69 \(⋮\) (x + 5)

=> (4x + 69) - (4x + 20) \(⋮\)(x + 5)

=> 49 \(⋮\) (x + 5)

=> x + 5 \(\in\) Ư(49) = {1; 7; 49}

Vì x \(\in\) N

=> x \(\in\) {2; 47}

c) Ta có: 2x + 1 \(⋮\) 2x + 1

=> 10x + 5 \(⋮\) (2x + 1)

Và 10x + 23 \(⋮\) (2x + 1)

=> (10x + 23) - (10x + 5) \(⋮\) (2x + 1)

=> 18 \(⋮\) (2x + 1)

=> 2x + 1 \(\in\) Ư(18) = {1; 2; 9; 18}

=> 2x \(\in\) {0; 1; 8; 17}

Vì x \(\in\) N

=> x \(\in\) {0; 4}