Question

chứng minh \(\dfrac{1}{3}\le\dfrac{a^2-2a+4}{a^2+2a+4}\le3\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a^2-2a+4}{a^2+2a+4}>=\dfrac{1}{3}\\\dfrac{a^2-2a+4}{a^2+2a+4}< =3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3a^2-6a+12-a^2-2a-4>=0\\a^2-2a+4-3a^2-6a-12< =0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2a^2-8a+8>=0\\-2a^2-8a-8< =0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\left(a-2\right)^2>=0\left(đúng\right)\\-2\left(a+2\right)^2< =0\left(đúng\right)\end{matrix}\right.\)

=>ĐPCM