Câu 1 (4,5 điểm). Cho biểu thức:
\[
A = \left( \frac{x^2 + 3x}{x^3 + 3x^2 + 9x + 27} + \frac{3}{x^2 + 9} \right) : \left( \frac{1}{(x-3)} - \frac{6x}{x^3 - 3x^2 + 9x - 27} \right) \, (\text{với } x \neq \pm 3)
\]
a) Rút gọn biểu thức \( A \);
b) Tìm số nguyên dương \( x \) để biểu thức \( A \) nhận giá trị nguyên;
c) Tìm \( x \) để \( A \ge 2 \)
a: \(A=\left(\dfrac{x^2+3x}{x^3+3x^2+9x+27}+\dfrac{3}{x^2+9}\right):\left(\dfrac{1}{x-3}-\dfrac{6x}{x^3-3x^2+9x-27}\right)\)
\(=\left(\dfrac{x\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}+\dfrac{3}{x^2+9}\right):\left(\dfrac{1}{x-3}-\dfrac{6x}{x^2\left(x-3\right)+9\left(x-3\right)}\right)\)
\(=\dfrac{x+3}{x^2+9}:\dfrac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}\)
\(=\dfrac{x+3}{x^2+9}\cdot\dfrac{\left(x-3\right)\left(x^2+9\right)}{\left(x-3\right)^2}=\dfrac{\left(x+3\right)}{x-3}\)
b: Để A là số nguyên thì \(x+3⋮x-3\)
=>\(x-3+6⋮x-3\)
=>\(6⋮x-3\)
=>\(x-3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
=>\(x\in\left\{4;2;5;1;6;0;9;-3\right\}\)
mà x là số nguyên dương và x<>3
nên \(x\in\left\{4;2;5;1;6;9\right\}\)
c: A>=2
=>A-2>=0
=>\(\dfrac{x+3}{x-3}-2>=0\)
=>\(\dfrac{x+3-2\left(x-3\right)}{x-3}>=0\)
=>\(\dfrac{-x+9}{x-3}>=0\)
=>\(\dfrac{x-9}{x-3}< =0\)
=>3<x<=9
a)
\(A=\left(\dfrac{x\left(x+3\right)}{\left(x^2+9\right)\left(x+3\right)}+\dfrac{3}{x^2+9}\right):\left(\dfrac{1}{x-3}-\dfrac{6x}{\left(x^2+9\right)\left(x-3\right)}\right)\)
\(A=\dfrac{x+3}{x^2+9}:\dfrac{x^2-6x+9}{\left(x-3\right)\left(x^2+9\right)}\)
\(A=\dfrac{x+3}{x^2+9}\cdot\dfrac{\left(x^2+9\right)\left(x-3\right)}{\left(x-3\right)^2}\)
\(A=\dfrac{x+3}{x-3}\)
b) Ta có \(A=\dfrac{x+3}{x-3}=\dfrac{\left(x-3\right)+6}{x-3}=1+\dfrac{6}{x-3}\)
Để biểu thức A nguyên thì \(\dfrac{6}{x-3}\) nguyên
\(\rightarrow6⋮x-3\rightarrow x-3\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
mà x là số nguyên dương nên \(x-3\ge-2\) và \(x\ne\pm3\rightarrow x-3\notin\left\{0;-6\right\}\)nên \(x-3\in\left\{\pm1;\pm2;3;6\right\}\)
\(\rightarrow x\in\left\{1;2;4;5;6;9\right\}\) (T/m)
c) \(A\ge2\rightarrow\dfrac{x+3}{x-3}\ge2\)
\(\rightarrow\dfrac{x+3}{x-3}-\dfrac{2\left(x-3\right)}{x-3}\ge0\)
\(\rightarrow\dfrac{-x+9}{x-3}\ge0\)
TH1:
\(\left\{{}\begin{matrix}-x+9\ge0\\x-3>0\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}x\le9\\x>3\end{matrix}\right.\rightarrow3< x\le9\)
\(\left\{{}\begin{matrix}-x+9\le0\\x-3< 0\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}x\ge9\\x< 3\end{matrix}\right.\left(L\right)\)
Vậy \(3< x\le9\) thì \(A\ge2\)
