a) ĐKXĐ: \(x\ne-3,x\ne2\)
b) \(D=\dfrac{\left(x+1\right)\left(x-2\right)-10+5\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2+4x+3}{\left(x+3\right)\left(x-2\right)}=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{x+1}{x-2}\)
c) \(D=\dfrac{x+1}{x-2}=\dfrac{x-2+3}{x-2}=1+\dfrac{3}{x-2}\in Z\)
\(\Leftrightarrow\left(x-2\right)\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
Kết hợp với ĐKXĐ: \(\Rightarrow x\in\left\{-1;1;3;5\right\}\)
b: \(D=\dfrac{\left(x+1\right)\left(x-2\right)-10+5\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-x-2-10+5x+15}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2+4x+3}{\left(x+3\right)\left(x-2\right)}=\dfrac{x+1}{x-2}\)
c: Để D là số nguyên thì \(x+1⋮x-2\)
\(\Leftrightarrow x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
a: ĐKXĐ: \(x\notin\left\{-3;2\right\}\)
a)D xác định <=>\(\left\{{}\begin{matrix}x+3\ne0\\x-2\ne0\end{matrix}\right.< =>\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)
b) Với x\(\ne-3,x\ne2\)
D=
\(\dfrac{x+1}{x+3}-\dfrac{10}{\left(x+3\right)\left(x-2\right)}+\dfrac{5}{x-2}=\dfrac{\left(x+1\right)\left(x-2\right)-10+5\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2-x-2-10+5x+15}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2+4x+3}{\left(x+3\right)\left(x-2\right)}=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{x+1}{x-2}\)
c)\(D=\dfrac{x+1}{x-2}=\dfrac{x-2+3}{x-2}=1+\dfrac{3}{x-2}\)
D có giá trị nguyên <=> 3\(⋮\)(x-2)
<=>x-2 là ước của 3
<=>x-2\(\varepsilon\left\{-3;-1;1;3\right\}\)
Ta có bảng
| x-2 | -3 | -1 | 1 | 3 |
| x | -1 | 1 | 3 | 5 |
| tm | tm | tm | tm |
