a, \(A=\frac{x-1}{x+1}=\frac{x+1-1-1}{x+1}=\frac{x+1-2}{x+1}=1-\frac{2}{x+1}\)
Để \(A\in z\) thì \(x+1\inƯ\left(2\right)=\left(-2;-1:1;2\right)\)
\(x+1=-2\Rightarrow x=-3\)
\(x+1=-1\Rightarrow x=-2\)
\(x+1=1\Rightarrow x=0\)
\(x+1=2\Rightarrow x=1\)
Vậy \(x=\left(-3;-2;0;1\right)\)thì \(A\in z\)
b, \(A=\frac{x+1}{x-2}=1+\frac{3}{x-2}\)
Để \(A\in z\)thì \(x-2\inƯ\left(3\right)=\left(-3;-1;1;3\right)\)
\(x-2=-3\Rightarrow x=-1\)
\(x-2=-1\Rightarrow x=1\)
\(x-2=1\Rightarrow x=3\)
\(x-2=3\Rightarrow x=5\)
Vậy \(x=\left(-1;1;3;5\right)\)thì \(A\in z\)
c, \(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)\(ĐK:\)\(x\ge0;x\ne9\)
\(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
Để \(A\in z\)thì \(\sqrt{x}-3\inƯ\left(4\right)=\left(-4;-2;-1;1;2;4\right)\)
\(\sqrt{x}-3=-4\Rightarrow\sqrt{x}=-1VN\)
\(\sqrt{x}-3=-2\Rightarrow\sqrt{x}=1\Rightarrow x=1\)
\(\sqrt{x}-3=-1\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
\(\sqrt{x}-3=1\Rightarrow\sqrt{x}=4\Rightarrow x=16\)
\(\sqrt{x}-3=2\Rightarrow\sqrt{x}=5\Rightarrow x=25\)
\(\sqrt{x}-3=4\Rightarrow\sqrt{x}=7\Rightarrow x=49\)
Vậy \(x=\left(1;4;16;25;49\right)\)thì \(A\in z\)
d, \(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}\) \(ĐK:\)\(x\ge0;x\ne1\)
\(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}=1+\frac{2}{\sqrt{x}-1}\)
Để \(A\in z\) thì \(\sqrt{x}-1\inƯ\left(2\right)=\left(-2;-1;1;2\right)\)
\(\sqrt{x}-1=-2\Rightarrow\sqrt{x}=-1VN\)
\(\sqrt{x}-1=-1\Rightarrow\sqrt{x}=0\Rightarrow x=0\)
\(\sqrt{x}-1=1\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
\(\sqrt{x}-1=2\Rightarrow\sqrt{x}=3\Rightarrow x=9\)
Vậy \(x=\left(0,4,9\right)\)thì \(A\in z\)
\(a,A=\frac{x-1}{x+1}\)
Để \(A\in Z\)
\(\Rightarrow\frac{x-1}{x+1}\in Z\)
\(\Rightarrow\frac{x+1-2}{x+1}\in Z\)
\(\Rightarrow1-\frac{2}{x+1}\in Z\)
\(\Rightarrow\frac{2}{x+1}\in Z\)
\(\Rightarrow x+1\in U_{\left(2\right)}\)
\(\Rightarrow x+1=\left\{-2,-1,1,2\right\}\)
\(\Rightarrow x=\left\{-3,-2,0,1\right\}\)
\(b;A=\frac{x+1}{x-2}=\frac{x-2+3}{x-2}=\frac{x-2}{x-2}+\frac{3}{x-2}=1+\frac{3}{x-2}\)
Để \(A\in Z\Rightarrow1+\frac{3}{x-2}\in Z\)
\(\Rightarrow\frac{3}{x-2}\in Z\)
\(\Rightarrow x-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Ta có bảng sau:
x-2 | -3 | -1 | 1 | 3 |
x | -1 | 1 | 3 | 5 |
\(\Rightarrow x\in\left\{-1;1;3,5\right\}\)