Question

tim x\(\in\)z

(4x+1)\(⋮\)(2x+2)

Answer 1

\(4x+1⋮2x+2\)

\(\Rightarrow4x+4-3⋮2x+2\)

\(\Rightarrow2\left(2x+2\right)-3⋮2x+2\)

      \(2\left(2x+2\right)⋮2x+2\)

\(\Rightarrow3⋮2x+2\)

\(\Rightarrow2x+2\inƯ\left(3\right)\)

\(\Rightarrow2x+2\in\left\{-1;1;-3;3\right\}\)

\(\Rightarrow2x\in\left\{-3;-1;-5;1\right\}\) 

\(\Rightarrow x\in\left\{-1,5;-0,5;-2,5;0,5\right\}\) mà x thuộc Z

\(\Rightarrow x\in\varnothing\)

Answer 2

\(\frac{4x+1}{2x+2}=\frac{2\left(2x+2\right)-3}{2x+2}=2-\frac{3}{2x+2}\)

\(\Rightarrow3⋮\left(2x+2\right)\Rightarrow2x+2\inƯ\left(3\right)=\left\{1;3;-1;-3\right\}\)

Nếu 2x+2=1 => 2x=-1 => x=-1/2

Nếu 2x+2=3 => 2x=1 => x=1/2

Nếu 2x+2=-1 => 2x=-3 => x=-3/2

Nếu 2x+2=-3 => 2x=-5 => x = -5/2

Vậy x = {-1/2 ; 1/2 ; -3/2 ; -5/2} thì 4x+1 chia hết cho 2x+2

Answer 3

4x+1\(⋮2x+2\)

\(\Rightarrow4x+4-3⋮2x+2\)

Do \(4x+4⋮2x+2\)

\(\Rightarrow3⋮2x+2\)

\(\Rightarrow2x+2\in\left(1;3;-1;-3\right)\)

Do 2x + 2 chẵn => ko có x thuộc Z thỏa mãn

Answer 4

\(\frac{4x+1}{2x+2}=\frac{2\left(2x+2\right)-3}{2x+2}=2-\frac{3}{2x+2}\)

\(\Rightarrow3⋮\left(2x+2\right)\Rightarrow2x+2\inƯ\left(3\right)=\left\{1;3;-1;-3\right\}\)

Nếu\(2x+2=1\Rightarrow2x=-1\Rightarrow x=\frac{-1}{2}\)

Nếu\(2x+2=3\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)

Nếu \(2x+2=1\Rightarrow2x=-3\Rightarrow x=\frac{-3}{2}\)

Nếu\(2x+2=-3\Rightarrow2x=-5\Rightarrow x=\frac{-5}{2}\)

Vậy\(x=\left\{\frac{-1}{2};\frac{1}{2};\frac{-3}{2};\frac{5}{2}\right\}\)thì \(4x+1\)chia hết cho \(2x+2\)