Question

\(log_2\left(log_{\frac{9}{16}}\left(x^2-4x+3\right)\right)< =0\)

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}x^2-4x+3>0\\log_{\frac{9}{16}}\left(x^2-4x+3\right)>0\end{matrix}\right.\)

\(\Rightarrow0< x^2-4x+3< 1\) \(\Rightarrow\left[{}\begin{matrix}2-\sqrt{2}< x< 1\\3< x< 2+\sqrt{2}\end{matrix}\right.\)

Khi đó:

\(\Leftrightarrow log_{\frac{9}{16}}\left(x^2-4x+3\right)\le1\)

\(\Leftrightarrow x^2-4x+3\ge\frac{9}{16}\)

\(\Leftrightarrow x^2-4x+\frac{39}{16}\ge0\Rightarrow\left[{}\begin{matrix}x\ge\frac{13}{4}\\x\le\frac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2-\sqrt{2}< x\le\frac{3}{4}\\\frac{13}{4}\le x< 2+\sqrt{2}\end{matrix}\right.\)