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Để A thuộc Z thì \(3x^2-x+1⋮3x+2\)
\(3x^2+2x-3x-2+3⋮3x+2\)
\(x\left(3x+2\right)-\left(3x+2\right)+3⋮3x+2\)
\(\left(3x+2\right)\left(x-1\right)+3⋮3x+2\)
Mà \(\left(3x+2\right)\left(x-1\right)⋮3x+2\)
\(\Rightarrow3⋮3x+2\)
\(\Rightarrow3x+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Ta có bảng :
| 3x+2 | 1 | -1 | 3 | -3 |
| x | -1/3 | -1 | 1/3 | -5/3 |
Mà x thuộc Z => x = -1
Vậy x = -1
\(A=\dfrac{3x^2+2x-3x+1}{3x+2}=\dfrac{3x^2+2x-3x-2+3}{3x+2}\)
\(A=\dfrac{x\left(3x+2\right)-\left(3x+2\right)+3}{3x+2}=x-1+\dfrac{3}{3x+2}\in Z\)
\(\Rightarrow3x+2\inƯ\left(3\right)\)
Xét ước thôi
a) \(A=\left(\dfrac{1}{3}+\dfrac{3}{x^2-3x}\right):\left(\dfrac{x^2}{27-3x^2}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3\left(x^2-3x\right)}:\left(\dfrac{x^2}{3\left(9-x^2\right)}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\left(\dfrac{x^2}{3.\left(3-x\right).\left(3+x\right)}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\dfrac{x^2+3.\left(3-x\right)}{3.\left(3-x\right).\left(3+x\right)}\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\dfrac{x^2+9-3x}{3.\left(3-x\right).\left(3+x\right)}\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}.\dfrac{3.\left(3x-x\right).\left(3+x\right)}{x^2+9-3x}\)
\(\Rightarrow A=\dfrac{1}{x.\left(x-3\right)}.\left(-\left(x-3\right)\right).\left(3+x\right)\)
\(\Rightarrow A=\dfrac{1}{x}.\left(-1\right).\left(3+x\right)\)
\(\Rightarrow A=-\dfrac{1}{x}.\left(3+x\right)\)
\(A=\left(\dfrac{1}{x-1}+\dfrac{x}{x^3-1}.\dfrac{x^2+x+1}{x+1}\right):\dfrac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x^2+x+1\right)}.\dfrac{x^2+x+1}{x+1}\right):\dfrac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}+\dfrac{x}{\left(x+1\right)\left(x-1\right)}\right).\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\dfrac{2x+1}{\left(x+1\right)\left(x-1\right)}.\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\dfrac{x+1}{x-1}\)
Vậy \(A=\dfrac{x+1}{x-1}\)
Giả sử tìm được \(x\in Z\) để \(A\in Z\)
\(x\in Z\Leftrightarrow\left\{{}\begin{matrix}x+1\in Z\\x-1\in Z\end{matrix}\right.\)
\(A=\dfrac{x+1}{x-1}=\dfrac{x-1+2}{x-1}=1+\dfrac{2}{x-1}\)
\(\Leftrightarrow2⋮x-1\Leftrightarrow x-1\inƯ\left(2\right)\)
Ta có các trường hợp :
+) \(x-1=1\Leftrightarrow x=2\)
+) \(x-1=2\Leftrightarrow x=3\)
+) \(x-1=-1\Leftrightarrow x=0\)
+) \(x-1=-2\Leftrightarrow x=-1\)
Vậy..
a: Để A nguyên thì \(x-3\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{4;2;5;1\right\}\)
b: Để B nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{3;1;5;-1\right\}\)
c: Để C nguyên thì \(3x^2+2x-3x-2+3⋮3x+2\)
=>\(3x+2\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{-\dfrac{1}{3};-1;\dfrac{1}{3};-\dfrac{5}{3}\right\}\)
Bài 1:
a: Để M>1 thì M-1>0
\(\Leftrightarrow\dfrac{x+4-x+4}{x-4}>0\)
=>x-4>0
hay x>4
Để M<2 thì M-2<0
\(\Leftrightarrow\dfrac{x+4-2x+8}{x-4}< 0\)
\(\Leftrightarrow-\dfrac{x-12}{x-4}< 0\)
=>x>12 hoặc x<4
b: Để M là số nguyên thì \(x-4+8⋮x-4\)
\(\Leftrightarrow x-4\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
hay \(x\in\left\{5;3;6;2;8;0;12;-4\right\}\)
a: ĐKXĐ: x<>1; x<>-1; x<>0
\(A=\dfrac{x}{3\left(x+1\right)}:\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x}{3\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{4x}=\dfrac{x-1}{12}\)
b: Để A là số nguyên thì x-1=12k
=>x=12k+1(k\(\in Z\))
a: \(A=\dfrac{2x-5+x^2-4+x^2-9}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2+2x-18}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x+6}{x-3}\)
b: Để A/2=x+3/x-3 là số nguyên thì \(x-3+6⋮x-3\)
=>\(x-3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
hay \(x\in\left\{4;51;6;0;9;-3\right\}\)
c: Để A=1/x-1 thì \(\dfrac{2x+6}{x-3}=\dfrac{1}{x-1}\)
=>2x^2-2x+6x-6=x-3
=>2x^2+5x-6-x+3=0
=>2x^2+4x-3=0
hay \(x=\dfrac{-2\pm\sqrt{10}}{2}\)
a/ Để A ∈ Z
⇒ \(3x^2-9x+2\) ⋮ \(x-3\)
⇒ \(3x\left(x-3\right)+2\) ⋮ \(x-3\)
Vì \(3x\left(x-3\right)\) ⋮ \(x-3\)
⇒ \(2\) ⋮ \(x-3\)
⇒ \(x-3\inƯ_{\left(2\right)}\)
⇒ \(x-3\in\left\{1;2;-1;-2\right\}\)
⇒ \(x\in\left\{4;5;2;1\right\}\)
Vậy ...
b.
Ta có:
\(A=\dfrac{3n+9}{n-4}=\dfrac{3\left(n-4\right)+21}{n-4}=3+\dfrac{21}{n-4}\)
Để A thuộc Z
=> \(\dfrac{21}{n-4}\in Z\) ( n khác 4)
=> \(21⋮\left(n-4\right)\)
\(\Rightarrow n-4\inƯ\left(21\right)=\left\{21;-21;7;-7;3;-3\right\}\)
\(\Rightarrow n\in\left\{25;-17;11;-3;-1;1\right\}\) ( nhận)