Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Do m, n cùng dấu, m, n khác 0 nên m, n cùng âm hoặc cùng dương, mà nếu m, n cùng âm thì \(\frac{1}{2m}+\frac{1}{n}< 0< \frac{1}{3}\)
trái với gt \(\Rightarrow\) m, n cùng dương
\(\frac{1}{3}=\frac{1}{2m}+\frac{1}{n}\ge2\sqrt{\frac{1}{2mn}}\)\(\Leftrightarrow\)\(\frac{1}{2mn}\le\frac{1}{36}\)\(\Leftrightarrow\)\(mn\ge18\)\(\Rightarrow\)\(B\ge18\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\frac{1}{2m}=\frac{1}{n}\\\frac{1}{2m}+\frac{1}{n}=\frac{1}{3}\end{cases}\Leftrightarrow\hept{\begin{cases}m=3\\n=6\end{cases}}}\)
\(\frac{m-4}{2-m^2}< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m-4>0\\2-m^2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}m-4< 0\\2-m^2>0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}m>4\\m< \sqrt{2}\end{matrix}\right.\\\left\{{}\begin{matrix}m< 4\\m>\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\sqrt{2}< m< 4\)(1)
\(\frac{m-2m^2}{4-2m^2}>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m-2m^2>0\\4-2m^2>0\end{matrix}\right.\\\left\{{}\begin{matrix}m-2m^2< 0\\m-2m^2< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m< \sqrt{2}\\m\left(1-2m\right)>0\end{matrix}\right.\\\left\{{}\begin{matrix}m>\sqrt{2}\\m\left(1-2m\right)< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m< \sqrt{2}\\\left[{}\begin{matrix}\left\{{}\begin{matrix}m>0\\1-2m>0\end{matrix}\right.\\\left\{{}\begin{matrix}m< 0\\1-2m< 0\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}m>\sqrt{2}\\\left[{}\begin{matrix}\left\{{}\begin{matrix}m>0\\1-2m< 0\end{matrix}\right.\\\left\{{}\begin{matrix}m< 0\\1-2m>0\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m< \sqrt{2}\\\left\{{}\begin{matrix}m>0\\m< \frac{1}{2}\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}m>\sqrt{2}\\m>\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}0< m< \frac{1}{2}\\m>\sqrt{2}\end{matrix}\right.\) (2)
\(\underrightarrow{\left(1\right)\left(2\right)}\) \(\sqrt{2}< m< \frac{1}{2}\)
Áp dụng hệ thức Vi-ét,ta có :
\(\hept{\begin{cases}x_1+x_2=\frac{m-1}{1}=m-1\\x_1x_2=\frac{2m-6}{1}=2m-6\end{cases}}\)
\(\frac{x_1}{x_2}+\frac{x_2}{x_1}=\frac{5}{2}\Leftrightarrow\frac{x_1^2+x_2^2}{x_1x_2}=\frac{5}{2}\)
\(\Leftrightarrow\frac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}=\frac{5}{2}\)
\(\Leftrightarrow\frac{\left(m-1\right)^2-2\left(2m-6\right)}{2m-6}=\frac{m^2-6m+13}{2m-6}=\frac{5}{2}\)
\(\Leftrightarrow2m^2-12m+26=10m-30\Leftrightarrow2m^2-22m+56=0\)
\(\Leftrightarrow\orbr{\begin{cases}m=4\\m=7\end{cases}}\)
Vây .....