Question

Bài 2: Giải pt, bpt sau:

6) \(\dfrac{2^{1-x} -2^x + 1}{2^x -1} ≤ 0\)

Answer 1

Đặt \(2^x=t>0\)

\(\Rightarrow\dfrac{\dfrac{2}{t}-t+1}{t-1}\le0\Leftrightarrow\dfrac{-t^2+t+2}{t\left(t-1\right)}\le0\)

\(\Leftrightarrow\dfrac{\left(2-t\right)\left(t+1\right)}{t\left(t-1\right)}\le0\)

\(\Leftrightarrow\dfrac{2-t}{t-1}\le0\) (do \(t>0;t+1>0\))

\(\Rightarrow\left[{}\begin{matrix}t\ge2\\t< 1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2^x\ge2\\2^x< 1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\ge1\\x< 0\end{matrix}\right.\)