Bài 6:
1: Để A có nghĩa thì \(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\end{matrix}\right.\)
=>\(x\in R\backslash\left\{2;-2\right\}\)
2: \(A=\left(\dfrac{3x}{x-2}-\dfrac{2x^2-5}{x^2-4}-\dfrac{x-1}{x+2}\right):\dfrac{3}{x+2}\)
\(=\dfrac{3x\left(x+2\right)-2x^2+5-\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{3}\)
\(=\dfrac{3x^2+6x-2x^2+5-\left(x^2-3x+2\right)}{x-2}\cdot\dfrac{1}{3}\)
\(=\dfrac{x^2+6x+5-x^2+3x-2}{x-2}\cdot\dfrac{1}{3}=\dfrac{9x+3}{3\left(x-2\right)}=\dfrac{3x+1}{x-2}\)
\(x^2-2x=0\)
=>x(x-2)=0
=>\(\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(loại\right)\end{matrix}\right.\)
Thay x=0 vào A, ta được:
\(A=\dfrac{3\cdot0+1}{0-2}=\dfrac{1}{-2}=-\dfrac{1}{2}\)
3: Để A nguyên thì \(3x+1⋮x-2\)
=>\(3x-6+7⋮x-2\)
=>\(7⋮x-2\)
=>\(x-2\in\left\{1;-1;7;-7\right\}\)
=>\(x\in\left\{3;1;9;-5\right\}\)
Bài 7:
1: \(A=\dfrac{x+1}{x-2}+\dfrac{x+1}{x+2}-\dfrac{x^2+4x}{4-x^2}\)
\(=\dfrac{\left(x+1\right)\left(x-2\right)+\left(x+1\right)\left(x+2\right)+x^2+4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-2+x+2\right)+x^2+4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+4x+2x^2+2x}{\left(x-2\right)\left(x+2\right)}=\dfrac{3x^2+6x}{\left(x-2\right)\left(x+2\right)}=\dfrac{3x}{x-2}\)
2: \(x^2-6x+8=0\)
=>(x-2)(x-4)=0
=>\(\left[{}\begin{matrix}x=2\left(loại\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
Thay x=4 vào A, ta được:
\(A=\dfrac{3\cdot4}{4-2}=\dfrac{12}{2}=6\)
3: Để A là số nguyên dương thì \(\left\{{}\begin{matrix}3x⋮x-2\\A>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x-6+6⋮x-2\\\dfrac{x}{x-2}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6⋮x-2\\\left[{}\begin{matrix}x>2\\x< 0\end{matrix}\right.\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\\\left[{}\begin{matrix}x>2\\x< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x\in\left\{3;4;5;-1;8;-4\right\}\)
