\(a,B=\left(\dfrac{3}{x^2+6x+9}-\dfrac{1}{x+3}\right):\left(\dfrac{3}{x^2-9}+\dfrac{1}{3-x}\right)\)
\(=\left(\dfrac{3}{\left(x+3\right)^2}-\dfrac{1}{x+3}\right):\left(\dfrac{3}{\left(x-3\right)\left(x+3\right)}-\dfrac{1}{x-3}\right)\)
\(=\dfrac{3-x-3}{\left(x+3\right)^2}:\dfrac{3-\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{-x}{\left(x+3\right)^2}.\dfrac{\left(x-3\right)\left(x+3\right)}{-x}\)
\(=\dfrac{x-3}{x+3}\left(dpcm\right)\)
\(b,x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Với \(x=2\Rightarrow B=\dfrac{2-3}{2+3}=-\dfrac{1}{5}\)
Với \(x=0\Rightarrow B=\dfrac{0-3}{0+3}=-\dfrac{3}{3}=-1\)
\(c,B\le-3\Leftrightarrow\dfrac{x-3}{x+3}\le-3\\ \Leftrightarrow\dfrac{x-3+3\left(x+3\right)}{x+3}\le0\\ \Leftrightarrow x-3+3x+9\le0\\ \Leftrightarrow4x+6\le0\\ \Leftrightarrow x\le-\dfrac{3}{2}\)
Mà x là giá trị nguyên nên \(x\le-2\)
Vậy \(x\le-2\) thì \(B\le-3\left(dk:x\in Z\right)\)
a)
\(B=\left(\dfrac{3}{x^2+6x+9}-\dfrac{1}{x+3}\right):\left(\dfrac{3}{x^2-9}+\dfrac{1}{3-x}\right)\\ =\left(\dfrac{3}{\left(x+3\right)^2}-\dfrac{x+3}{\left(x+3\right)^2}\right):\left(\dfrac{3}{\left(x-3\right)\left(x+3\right)}-\dfrac{1}{x-3}\right)\)
\(=\dfrac{3-x-3}{\left(x+3\right)^2}:\left(\dfrac{3}{\left(x-3\right)\left(x+3\right)}-\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\right)\)
\(=\dfrac{-x}{\left(x+3\right)^2}:\dfrac{3-x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{-x}{\left(x+3\right)^2}\cdot\dfrac{\left(x-3\right)\left(x+3\right)}{-x}\\ =\dfrac{x-3}{x+3}\)
b)
`x^2 -2x=0`
`<=>x(x-2)=0`
`<=>x=0` hoặc `x-2=0`
`<=>x=0(ktm)` hoặc `x=2(tm)`
Với x=2 thì
\(\dfrac{2-3}{2+3}=\dfrac{-1}{5}\)
c)
Dể `B<=-3`
`=>(x-3)/(x+3) <=-3`
`<=>(x-3)/(x+3)+3 <=0`
`<=>(x-3)/(x+3)+(3(x+3))/(x+3) <=0`
`<=>(x-3+3x+9)/(x+3) <=0`
