Question

Giải bất phương trình: \(\left|\frac{x^2-5x+4}{x^2-4}\right|\le1\)

Answer 1

\(x\ne\pm2\)

\(\Leftrightarrow\left(\frac{x^2-5x+4}{x^2-4}\right)^2\le1\)

\(\Leftrightarrow\left(x^2-5x+4\right)^2\le\left(x^2-4\right)^2\)

\(\Leftrightarrow\left(x^2-5x+4\right)^2-\left(x^2-4\right)^2\le0\)

\(\Leftrightarrow\left(x^2-5x+4-x^2+4\right)\left(x^2-5x+4+x^2-4\right)\le0\)

\(\Leftrightarrow\left(8-5x\right)\left(2x^2-5x\right)\le0\)

\(\Leftrightarrow x\left(8-5x\right)\left(2x-5\right)\le0\)

\(\Rightarrow\left[{}\begin{matrix}0\le x\le\frac{8}{5}\\x\ge\frac{5}{2}\end{matrix}\right.\)