Question

tìm x biết x là số nguyên

(10x+9) chia hết cho (2x+3)

(4x+1) chia hết cho (2x-1)

mình cần gấp giúp nha

Answer 1

1) Ta có: \(10x+9⋮2x+3\)

\(\Leftrightarrow10x+15-6⋮2x+3\)

\(\Leftrightarrow5\left(2x+3\right)-6⋮2x+3\)

Do 5(2x+3)\(⋮\) 2x+3 \(\Rightarrow6⋮2x+3\)

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

\(\Rightarrow2x\in\left\{-2;-4;0;-6\right\}\)

\(\Rightarrow x\in\left\{-1;-2;0;-3\right\}\)

Vậy \(x\in\left\{-1;-2;0;-3\right\}\)

2) Ta có: \(4x+1⋮2x-1\)

\(\Leftrightarrow4x-2+3⋮2x-1\)

\(\Leftrightarrow2\left(2x-1\right)+3⋮2x-1\)

Do 2(2x-1) \(⋮\) 2x-1 \(\Rightarrow3⋮2x-1\)

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

\(\Rightarrow2x\in\left\{2;0;4;-2\right\}\)

\(\Rightarrow x\in\left\{1;0;2;-1\right\}\)

Vậy:.............