a, Để phân số đạt giá trị nguyễn
\(\Rightarrow x+1⋮x-2\)
\(\Rightarrow x-2+3⋮x-2\)
mà \(x-2⋮x-2\Rightarrow3⋮x-2\)
\(\Rightarrow x-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow x\in\left\{3;5\pm1\right\}\)
b,Tương tự :
\(2x-1⋮x+5\)
\(\Rightarrow2x+10-11⋮x+5\)
\(2\left(x+5\right)-11⋮x+5\)
mà \(2\left(x+5\right)⋮x+5\Rightarrow11⋮x+5\)
\(\Rightarrow x+5\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)
\(x\in\left\{-4;\pm6;-16\right\}\)
a, Để \(A\in Z\)\(\Leftrightarrow x+1⋮x-2\)\
Ta có: \(\hept{\begin{cases}x-2⋮x-2\\x+1⋮x-2^{ }_{ }\end{cases}}\) \(\Rightarrow\) \(x-2-\left(x+1\right)⋮x-2\)
\(\Rightarrow-3⋮x-2\)mà \(x\in Z\Rightarrow x-2\in Z\Rightarrow x-2\inƯ\left(-3\right)\)\(\in\left(1;-1,3;-3\right)\)
\(x\in\left(3;1;5;-1\right)\)Vậy: \(x\in\left(1;3;5;-1\right)\)thì \(A\in Z\)
b, Để \(B\in Z\Leftrightarrow2x-1⋮x+5\)
Mà \(\hept{\begin{cases}x+5⋮x+5\\2x-1⋮x+5\end{cases}}\)\(\Rightarrow\)\(2x+10-\left(2x-1\right)⋮x+5\)\(\Rightarrow11⋮x+5\)
Vì \(x\in Z\Rightarrow x+5\in Z\Rightarrow x+5\inƯ\left(11\right)\)\(\in\left(1;11;-1;-11\right)\)
\(\Rightarrow x\in\left(-4;6;-6;-16\right)\)
Vậy : ...........