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a) \(A=\dfrac{x+3}{x+2}=\dfrac{x-2+5}{x-2}=\dfrac{x-2}{x-2}+\dfrac{5}{x-2}=1+\dfrac{5}{x-2}\)
\(\Rightarrow5⋮x-2\Rightarrow x-2\inƯ\left(5\right)\)
\(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\\x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=1\\x=7\\x=-3\end{matrix}\right.\)
b) \(B=\dfrac{1-2x}{x+3}=\dfrac{-2x+1}{x+3}\)
\(B\in Z\Rightarrow-2x+1⋮x+3\)
\(\Rightarrow-2x-6+7⋮x+3\)
\(\Rightarrow-2\left(x+3\right)+7⋮x+3\)
\(\Rightarrow7⋮x+3\)
\(\Rightarrow x+3\inƯ\left(7\right)\)
\(Ư\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow\left[{}\begin{matrix}x+3=1\\x+3-1\\x+3=7\\x+3=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=-4\\x=4\\x=-10\end{matrix}\right.\)
\(A=\dfrac{x+3}{x-2}=\dfrac{x-2+5}{x-2}=1+\dfrac{5}{x-2}\)
Để \(A\in Z\) thì \(x-2\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
\(\Rightarrow x\in\left\{3;1;7;-3\right\}\)
Vậy \(x\in\left\{3;1;7;-3\right\}\) thì \(A\in Z\)
\(B=\dfrac{1-2x}{x+3}=\dfrac{-2x-6+7}{x+3}=\dfrac{-2\left(x+3\right)-7}{x+3}=-2+\dfrac{-7}{x+3}\)
Để \(B\in Z\) thì \(x+3\inƯ\left(-7\right)=\left\{1;-1;7;-7\right\}\)
\(\Rightarrow x\in\left\{-2;-4;4;10\right\}\)
Vậy \(x\in\left\{-2;-4;4;10\right\}\) thì \(B\in Z\)
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A=7/x+3 -2 để A thuộc Z thì x+3 là ước của 7.
=>x+3=(+1,-1;+7,-7)
x=-2 =>A=5 x=4=>A=-1
x=-4=> A=-9 x=-10=>A=-3
\(B=\frac{1-2x}{x+3}=\frac{1-2x-6+6}{x+3}=\frac{7-2\left(x+3\right)}{x+3}=\frac{7}{x+3}-2\)
Để \(A=\frac{7}{x+3}-2\) là số nguyên <=> \(\frac{7}{x+3}\) là số nguyên
=> x + 3 thuộc Ư(7) = { - 7; - 1; 1; 7 }
+ ) Với x + 3 = - 7 thì x = - 10 (TM)
+ ) Với x + 3 = - 1 thì x = - 4 (TM)
+ ) Với x + 3 = 1 thì x = - 2 (TM)
+ ) Với x + 3 = 7 thì x = 4 (TM)
Vậy x = { - 10; - 4; - 2; 4 }
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Để x thuộc Z thì : à dấu " : " là " chia hết cho " nhá ^^
2x - 1 : x + 2
2x + 2 -3 : x + 2
mà 2x + 2 : x + 2 => 3 : x + 2 => x + 2 thuộc Ư(3) = { 1; -1; 3; -3 }
+) x + 2 = 1
x = -1
+) x + 2 = -1
x = -3
+) x + 2 = 3
x = 1
+) x + 2 = -3
x = -5
Vậy,.........
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\(A=\frac{1-2x}{x+3}=\frac{-2x+1}{x+3}=\frac{-2x-6+7}{x+3}=\frac{-2\left(x+3\right)+7}{x+3}=-2+\frac{7}{x+3}\)
Vì \(-2\inℤ\)\(\Rightarrow\)Để \(A\inℤ\)thì \(\frac{7}{x+3}\inℤ\)
\(\Rightarrow7⋮x+3\)\(\Rightarrow x+3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow x\in\left\{-10;-4;-2;4\right\}\)
Vậy \(x\in\left\{-10;-4;-2;4\right\}\)
ĐK:\(x\ne-3\)
Với \(A=\frac{1-2X}{X+3}=\frac{-2x-6+7}{x+3}=\frac{-2+7}{x+3}\)
A nguyên <=>\(x+3\inƯ\left(7\right)\)\(\Rightarrow x\in\left\{1;-1;7;-7\right\}\)
Vậy...
A thuộc Z <=> 2x chia hết cho x-1
<=> 2(x-1) =2x-2 chia hết cho x-1
<=> 2x - (2x-2) chia hết cho x-1
<=> 2 chia hết cho x-1
<=> x - 1 \(\in\) {1;2;-1;-2}
<=> x \(\in\) {2;3;0;-1}
\(\frac{21}{4}\)