Question

Giải bpt \(\dfrac{log_2\dfrac{x}{2}}{log_2x}-\dfrac{log_2x^2}{log_2x-1}\le1\)

Answer 1

ĐKXĐ: \(x>0;x\ne\left\{1;2\right\}\)

\(\dfrac{log_2x-1}{log_2x}-\dfrac{2log_2x}{log_2x-1}\le1\)

Đặt \(log_2x=t\)

\(\Rightarrow\dfrac{t-1}{t}-\dfrac{2t}{t-1}\le1\)

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

\(\Rightarrow\left[{}\begin{matrix}t\le-1\\0< t\le\dfrac{1}{2}\\t>1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}log_2x\le-1\\0< log_2x\le\dfrac{1}{2}\\log_2x>1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\le\dfrac{1}{2}\\1< x\le\sqrt{2}\\x>2\end{matrix}\right.\)