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\(B=\dfrac{\left(a+3\right)^2}{2a^2+6a}\cdot\dfrac{1-6a-18}{a^2-9}\\ a,ĐK:a\ne0;a\ne\pm3\\ b,B=\dfrac{\left(a+3\right)^2}{2a\left(a+3\right)}\cdot\dfrac{-17-6a}{\left(a-3\right)\left(a+3\right)}=\dfrac{-17-6a}{2a\left(a-3\right)}\\ c,B=0\Leftrightarrow-17-6a=0\Leftrightarrow a=-\dfrac{17}{6}\left(tm\right)\\ d,B=1\Leftrightarrow-17-6a=2a^2-6a\\ \Leftrightarrow2a^2=-17\Leftrightarrow a\in\varnothing\)
a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
b) Ta có: \(B=\left(\dfrac{x-2}{2x-2}+\dfrac{3}{2x-2}-\dfrac{x+3}{2x+2}\right):\left(1-\dfrac{x-3}{x+1}\right)\)
\(=\left(\dfrac{x-1}{2x-2}-\dfrac{x+3}{2x+2}\right):\left(\dfrac{x+1-x-3}{x+1}\right)\)
\(=\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right):\dfrac{-2}{x+1}\)
\(=\dfrac{x^2-1-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{-2}\)
\(=\dfrac{-2x+2}{2\left(x-1\right)}\cdot\dfrac{-1}{2}\)
\(=\dfrac{-2\left(x-1\right)}{2\left(x-1\right)}\cdot\dfrac{-1}{2}\)
\(=\dfrac{1}{2}\)
Vậy: Khi x=2005 thì \(B=\dfrac{1}{2}\)
A∈Z⇒\(\dfrac{2\left(x+1\right)}{x+3}\in Z\Rightarrow\left(2x+2\right)⋮\left(x+3\right)\)
\(\Rightarrow\left(2x+6-4\right)⋮\left(x+3\right)\\ \Rightarrow\left[2\left(x+3\right)-4\right]⋮\left(x+3\right)\)
\(\text{Mà}2\left(x+3\right)⋮\left(x+3\right)\\ \Rightarrow-4⋮\left(x+3\right)\\ \Rightarrow x+3\inƯ\left(-4\right)=\left\{-4;-2;-1;1;2;4\right\}\\ \Rightarrow x\in\left\{-7;-5;-4;-2;-1;1\right\}\)
