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12 tháng 5 2022
\(A=\dfrac{3x^2-9x+x-3+2}{x-3}\)
\(B=\dfrac{x^2\left(x+2\right)+5\left(x+2\right)}{\left(x+2\right)^2}=\dfrac{x^2+5}{x+2}=x-2+\dfrac{9}{x+2}\)
Để A và B cùng là số nguyên thì
\(\left\{{}\begin{matrix}x-3\in\left\{1;-1;2;-2\right\}\\x+2\in\left\{1;-1;3;-3;9;-9\right\}\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x\in\left\{4;2;5;1\right\}\\x\in\left\{-1;-3;1;-5;7;-11\right\}\end{matrix}\right.\)
hay x=1
B
8 tháng 12 2016
a) \(A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{x^2-3x+2}\)
\(\Leftrightarrow A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{x^2-x-2x+2}\)
\(\Leftrightarrow A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{x\left(x-1\right)-2\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{\left(x-1\right)\left(x-2\right)}\)
\(\Leftrightarrow A=\frac{\left(4x-1\right)\left(x-1\right)-\left(x-3\right)\left(x-2\right)-2x+4}{\left(x-2\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{4x^2-4x-x+1-x^2+2x+3x-6-2x+4}{\left(x-2\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{3x^2-2x-1}{\left(x-2\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{3x^2-3x+\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}\)\(=\frac{3x\left(x-1\right)+\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}\)\(=\frac{\left(x-1\right)\left(3x+1\right)}{\left(x-2\right)\left(x-1\right)}\)\(=\frac{3x+1}{x-2}\)
b)\(\frac{3x+1}{x-2}=\frac{3x-6+7}{x-2}=\frac{3x-6}{x-2}+\frac{7}{x-2}=3+\frac{7}{x-2}\)
Ta có : \(x-2\inƯ_7\left\{-7;-1;1;7\right\}\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-2=-7\\x-2=-1\\x-2=1\\x-2=7\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}\text{x=-5}\\\text{x=1}\\\text{x=3}\\\text{x}=9\end{array}\right.\)
\(\text{x}=1\) (loại)
Vậy giá trị nguyên tập hợp x là:
x=-5;3;9
10 tháng 6 2018
a) ta có: \(A=\frac{2x}{x-2}=\frac{2x-4+4}{x-2}=\frac{2.\left(x-2\right)+4}{x-2}=\frac{2.\left(x-2\right)}{x-2}+\frac{4}{x-2}=2+\frac{4}{x-2}\)
Để \(A\inℤ\)
\(\Rightarrow\frac{4}{x-2}\inℤ\)
\(\Rightarrow4⋮x-2\Rightarrow x-2\inƯ_{\left(4\right)}=\left(4;-4;2;-2;1;-1\right)\)
nếu x -2 = 4 => x = 6 (TM)
x- 2= - 4 => x= - 2 (TM)
x- 2= 2 => x = 4 (TM)
x- 2 = -2 => x = 0 (TM)
x - 2 = 1 => x = 3 (TM)
x - 2 = -1 => x= 1 (TM)
KL: \(x\in\left(6;-2;4;0;3;1\right)\)
c) ta có: \(C=\frac{x^2+2}{x+1}=\frac{\left(x+1\right).\left(x-1\right)+3}{x+1}=\frac{\left(x+1\right).\left(x-1\right)}{x+1}+\frac{3}{x+1}\)\(=x-1+\frac{3}{x+1}\)
Để \(C\inℤ\)
\(\Rightarrow\frac{3}{x+1}\inℤ\)
\(\Rightarrow3⋮x+1\Rightarrow x+1\inƯ_{\left(3\right)}=\left(3;-3;1;-1\right)\)
nếu x + 1 = 3 => x = 2 (TM)
x + 1 = - 3 => x = -4 (TM)
x + 1 = 1 => x = 0
x + 1 = -1 => x = -2 (TM)
KL: \(x\in\left(2;-4;0;-2\right)\)
p/s
Ta có:
\(B=\frac{2x^3+x^2+2x+4}{2x+1}=\frac{x^2.\left(2x+1\right)+2x+1+3}{2x+1}\)
\(B=\frac{\left(2x+1\right).\left(x^2+1\right)+3}{2x+1}\)
\(B=\frac{\left(2x+1\right).\left(x^2+1\right)}{2x+1}+\frac{3}{2x+1}\)
\(B=x^2+1+\frac{3}{2x+1}\)
Do x nguyên nên x2 + 1 nguyên
Để B nguyên thì \(\frac{3}{2x+1}\) nguyên
\(\Rightarrow3⋮2x+1\)
\(\Rightarrow2x+1\in\left\{1;-1;3;-3\right\}\)
\(\Rightarrow2x\in\left\{0;-2;2;-4\right\}\)
\(\Rightarrow x\in\left\{0;-1;1;-2\right\}\)
Vậy \(x\in\left\{0;-1;1;-2\right\}\)