Question

Bài 1: Cho biểu thức \(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}-\dfrac{1}{x-2}\)

a, Rút gọn biểu thức A

b, Tìm x biết A = -3

c, Tìm x nguyên để A đạt giá trị nguyên dương

Answer 1

\(a,\)Với \(x\ne-3,x\ne2\) ta có :

\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}-\dfrac{1}{x-2}\)

   \(=\dfrac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)

   \(=\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)

   \(=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)

   \(=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)

  \(=\dfrac{x-4}{x-2}\)

\(b,\) \(A=-3\Leftrightarrow\dfrac{x-4}{x-2}=-3\)

\(\Leftrightarrow x-4=-3\left(x-2\right)\)

\(\Leftrightarrow x-4+3x-6=0\)

\(\Leftrightarrow4x=10\Rightarrow x=\dfrac{10}{4}=\dfrac{5}{2}\)

Answer 2

c) Để A đạt giá trị nguyên dương thì \(\left\{{}\begin{matrix}x-4⋮x-2\\x-2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2⋮x-2\\x>2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2\inƯ\left(-2\right)\\x>2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2\in\left\{1;-1;2;-2\right\}\\x>2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{3;1;4;0\right\}\\x>2\end{matrix}\right.\Leftrightarrow x=4\)

Vậy: Để A là số nguyên dương thì x=4