Question

Biết B=\(\dfrac{x+2}{3-x}\) .Tìm x để \(B\le0\)

Answer 1

Để B \(\le\) 0 => \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2\le0\\3-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+2\ge0\\3-x< 0\end{matrix}\right.\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le-2\\x< 3\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge-2\\x>3\end{matrix}\right.\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x\le-2\\x>3\end{matrix}\right.\)

Answer 2

\(B\le0\)\(\Rightarrow\dfrac{x+2}{3-x}\le0\)

\(\Rightarrow\left\{{}\begin{matrix}x+2\ge0\\3-x< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge-2\\x>3\end{matrix}\right.\)

vì -2<3 nên \(x>3\)

vậy để B\(\le\)0 thì x>3