Ta có: \(4x+10⋮2x+3\)
\(\Rightarrow4x+6+4⋮2x+3\)
\(\Rightarrow4x+6⋮2x+3\)
\(\Rightarrow4⋮2x+3\)
\(\Rightarrow2x+3\inƯ_{\left(4\right)}=-4;-2;-1;1;2;4\)
* Nếu: \(2x+3=-4\Rightarrow2x=-7\Rightarrow x=\frac{-7}{2}\)
* Nếu: \(2x+3=-2\Rightarrow2x=-5\Rightarrow x=\frac{-5}{2}\)
* Nếu: \(2x+3=-1\Rightarrow2x=-4\Rightarrow x=-2\)
* Nếu: \(2x+3=1\Rightarrow2x=-2\Rightarrow x=-1\)
* Nếu: \(2x+3=2\Rightarrow2x=-1\Rightarrow x=-\frac{1}{2}\)
* Nếu: \(2x+3=4\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)
Vậy: \(x=-\frac{7}{2};-\frac{5}{2};-2;-1;-\frac{1}{2};\frac{1}{2}\)
Ta có: \(2x+6⋮x+2\)
\(\Rightarrow2x+4+2⋮x+2\)
\(\Rightarrow2x+4⋮x+2\)
\(\Rightarrow2⋮x+2\)
\(\Rightarrow x+2\inƯ_{\left(2\right)}=-2;-1;1;2\)
\(\Rightarrow\left[\begin{matrix}x+2=-2\\x+2=-1\\x+2=1\\x+2=2\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=-4\\x=-3\\x=-1\\x=0\end{matrix}\right.\)
Vậy: \(x=-4;-3;-1;0\)
có 2x+6\(⋮\)x+2
vì x+2\(⋮\)x+2=>2(x+2)\(⋮\)x+2
mà2x+6\(⋮\)x+2
=>2x+6-2x+4\(⋮\)x+2
=>2\(⋮\)x+2
=>x+2\(\in\)Ư(2)
=>x+2\(\in\)\(\left\{1;2\right\}\)
vậy x+2\(\in\)\(\left\{1;2\right\}\)
từ đó bn lm nốt các câu còn lại nhé
