b) x-7\(⋮\)x+6
\(\Rightarrow\)x+6-13\(⋮\)x+6
\(\Rightarrow\)13\(⋮\)x+6
\(\Rightarrow\)x+6 \(\in\) Ư(13)
Ta có Ư(13)={\(\pm1;\pm13\)}
nếu x+6=1 thì x=-5
nếu x+6=-1 thì x=-7
nếu x+6=13 thì x=7
nếu x+6=-13 thì x=-19
Vậy x \(\in\) {-5;-7;7;-19}
a) Ta có: \(-2x-11⋮3x+2\)
\(\Rightarrow-3\left(-2x-11\right)⋮3x+2\)
\(\Rightarrow6x+33⋮3x+2\)
\(\Rightarrow\left(6x+4\right)+29⋮3x+2\)
\(\Rightarrow2\left(3x+4\right)+29⋮3x+2\)
\(\Rightarrow29⋮3x+2\)
\(\Rightarrow3x+2\in\left\{1;-1;29;-29\right\}\)
\(\left[\begin{matrix}3x+2=1\\3x+2=-1\\3x+2=29\\3x+2=-29\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=-\frac{1}{3}\\x=-1\\x=9\\x=-\frac{31}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{-1}{3};-1;9;\frac{-31}{3}\right\}\)
b) \(x-7⋮x+6\)
\(\Rightarrow\left(x+6\right)-13⋮x+6\)
\(\Rightarrow13⋮x+6\)
\(\Rightarrow x+6\in\left\{1;-1;13;-13\right\}\)
\(\left[\begin{matrix}x+6=1\\x+6=-1\\x+6=13\\x+6=-13\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=-5\\x=-7\\x=7\\x=-19\end{matrix}\right.\)
Vậy \(x\in\left\{-5;-7;7;-19\right\}\)
a)-2x-11\(⋮\)3x+2
\(\Rightarrow\)-3(-2x-11)\(⋮\)3x+2
\(\Rightarrow\)6x+33\(⋮\)3x+2
\(\Rightarrow\)(6x+4)+29\(⋮\)3x+2
\(\Rightarrow\)2(3x+4)+29\(⋮\)3x+2
\(\Rightarrow\)29\(⋮\)3x+2
\(\Rightarrow\)3x+2 \(\in\) Ư(29)={\(\pm1;\pm29\)}
nếu 3x+2=1 thì x=\(\frac{-1}{3}\)
nếu 3x+2=-1 thì x=-1
nếu 3x+2=29 thì x=9
nếu 3x+2=-29 thì x=\(\frac{-31}{3}\)
Vậy x \(\in\) {\(\frac{-1}{3};-1;9;\frac{-31}{3}\)}
