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ĐKXĐ: \(x\ne\pm3\)
a
Khi x = 1:
\(A=\dfrac{3.1+2}{1-3}=\dfrac{5}{-2}=-2,5\)
Khi x = 2:
\(A=\dfrac{3.2+2}{2-3}=-8\)
Khi x = \(\dfrac{5}{2}:\)
\(A=\dfrac{3.2,5+2}{2,5-3}=\dfrac{9,5}{-0,5}=-19\)
b
Để A nguyên => \(\dfrac{3x+2}{x-3}\) nguyên
\(\Leftrightarrow3x+2⋮\left(x-3\right)\\3\left(x-3\right)+11⋮\left(x-3\right) \)
Vì \(3\left(x-3\right)⋮\left(x-3\right)\) nên \(11⋮\left(x-3\right)\)
\(\Rightarrow\left(x-3\right)\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\\ \Rightarrow x\left\{4;2;-8;14\right\}\)
c
Để B nguyên => \(\dfrac{x^2+3x-7}{x+3}\) nguyên
\(\Rightarrow x\left(x+3\right)-7⋮\left(x+3\right)\)
\(\Rightarrow-7⋮\left(x+3\right)\\ \Rightarrow x+3\inƯ\left\{\pm1;\pm7\right\}\)
\(\Rightarrow x=\left\{-4;-11;-2;4\right\}\)
d
\(\left\{{}\begin{matrix}A.nguyên.\Leftrightarrow x=\left\{-8;2;4;14\right\}\\B.nguyên\Leftrightarrow x=\left\{-11;-4;-2;4\right\}\end{matrix}\right.\)
=> Để A, B cùng là số nguyên thì x = 4.
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(B\in Z\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 }
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Để B nguyên thì \(-7⋮x+3\)
\(\Leftrightarrow x+3\in\left\{1;-1;7;-7\right\}\)
hay \(x\in\left\{-2;-4;4;-10\right\}\)
b: Để A là số nguyên thì \(3x+2⋮x-3\)
\(\Leftrightarrow x-3\in\left\{1;-1;11;-11\right\}\)
hay \(x\in\left\{-2;-4;14;-8\right\}\)
Để A và B cùng là số nguyên thì \(x\in\left\{-2;-4\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`B` thuộc `ZZ <=> (x+1) vdots (x-3)`
`<=>(x-3)+4 vdots (x-3)`
`<=> 4 vdots (x-3)`
`<=> x-3 \in {-4;4;-2;2;-1;1}`
`<=> x \in {-1;7;1;5;2;4}`