Question

C =\(\dfrac{x-2}{x+1}\)

a,Tìm x khi C > 0

b,Tìm x khi C < 1

c,Tìm x khi C = \(\dfrac{1}{4}\)

Answer 1

\(C>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -1\end{matrix}\right.\)

Vậy .......

Answer 2

b) Ta có: C<1

nên C-1<0

\(\Leftrightarrow\dfrac{x-2}{x+1}-1< 0\)

\(\Leftrightarrow\dfrac{x-2-x-1}{x+1}< 0\)

\(\Leftrightarrow\dfrac{-3}{x+1}< 0\)

\(\Leftrightarrow x+1>0\)

hay x>-1

c) Để \(C=\dfrac{1}{4}\) thì \(\dfrac{x-2}{x+1}=\dfrac{1}{4}\)

\(\Leftrightarrow4x-8-x-1=0\)

\(\Leftrightarrow3x=9\)

hay x=3

Answer 3

Ta có: C>0

nên \(\left[{}\begin{matrix}x-2>0\\x+1< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -1\end{matrix}\right.\)