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\(A=\frac{3x-1}{x-1}=\frac{3\left(x-1\right)+2}{x-1}=3+\frac{2}{x-1}\)
\(B=\frac{2x^2+x-1}{x+2}=\frac{\left(x+2\right)\left(2x-3\right)+5}{x+2}=2x-3+\frac{5}{x+2}\)
Để A,B đều là số nguyên thì \(x-1\in\left\{1;2;-1;-2\right\}\) và \(x+2\in\left\{1;5;-1;-5\right\}\)
Bạn tự làm nốt
\(\dfrac{x}{9}< \dfrac{4}{7}< \dfrac{x+1}{9}\)
=>\(\dfrac{7x}{63}< \dfrac{36}{63}< \dfrac{7x+7}{63}\)
\(\Rightarrow7x< 36< 7x+7\)
\(\Rightarrow x< \dfrac{36}{7}< x+1\)
\(\Rightarrow x< 5\dfrac{1}{7}< x+1\)
\(\Rightarrow x=5\)
\(\dfrac{x}{9}\) < \(\dfrac{4}{7}\) < \(x\) + \(\dfrac{1}{9}\)
\(\dfrac{7x}{63}\) < \(\dfrac{36}{63}\) < \(\dfrac{63x}{63}\) + \(\dfrac{7}{63}\)
7\(x\) < 36 < 63\(x\) + 7
⇒\(\left\{{}\begin{matrix}7x< 36\\63x+7>36\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>36-7\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>29\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\x>\dfrac{29}{63}\end{matrix}\right.\)
\(\dfrac{29}{63}\)< \(x\) < \(\dfrac{36}{7}\) vì \(x\in\) Z nên \(x\in\) { 1; 2; 3; 4; 5}
⇒ \(\dfrac{x}{9}\) = \(\dfrac{1}{9}\); \(\dfrac{2}{9}\); \(\dfrac{3}{9}\); \(\dfrac{4}{9}\);\(\dfrac{5}{9}\)
a, Với x = 1 ta có :
\(A=\frac{3.1+2}{1-3}=\frac{3+2}{1-3}=\frac{5}{-2}=-\frac{5}{2}\)
Với x = 2 ta có
\(A=\frac{3.2+2}{2-3}=\frac{6+2}{-1}=-8\)
Với x = \(\frac{5}{2}\)
\(A=\frac{3.\frac{5}{2}+2}{\frac{5}{2}-3}=\frac{\frac{15}{2}+2}{-\frac{1}{2}}=-\frac{\frac{19}{2}}{\frac{1}{2}}=-19\)
b, \(A=\frac{3x+2}{x-3}=3+\frac{11}{x-3}\) ĐK \(x\ne3\)
Để A nguyên \(\Rightarrow x-3\inƯ\left(11\right)=\left(-11;-1;1;11\right)\)
\(x-3=-11\Rightarrow x=-8\)
\(x-3=-1\Rightarrow x=2\)
\(x-3=1\Rightarrow x=4\)
\(x-3=11\Rightarrow x=14\)
Vậy \(x=\left(-8;2;4;14\right)\)thì A nguyên
ta có :A=8 bé hơn hoặc =|X-1+X-3+X-5+X-7|=|4X-16|
=>X<(8+16)/4=6(1)
X>(-8+16)/4=2(2)TỪ 1 VÀ 2 =>2<x<6
=>x\(\in\)(3;4;5)
\(x=\dfrac{5}{a-1}< 0\)
\(\Leftrightarrow a-1< 0\Leftrightarrow a< 1\left(1\right)\)
Và \(x=\dfrac{5}{a-1}\in Z\)
\(\Rightarrow a-1\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\left(2\right)\)
\(\Rightarrow a\left\{2;0;6;-4\right\}\)
\(\left(1\right),\left(2\right)\Rightarrow a\in\left\{-4;0\right\}\)