giải tiếp đc ko bạn :)

bạn thử đi, nhân chéo rồi thu gọn

mk chắc chắn 100% là 99m<9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999

ĐKXĐ: ...

Đặt \(x+\frac{1}{x}=a\Rightarrow x^2+\frac{1}{x^2}=a^2-2\) (với \(\left|a\right|\ge2\))

Phương trình trở thành:

\(a^2-2-2ma+2m+1=0\Leftrightarrow a^2-2ma+2m-1=0\)

\(\Leftrightarrow\left(a-1\right)\left(a+1\right)-2m\left(a-1\right)=0\)

\(\Leftrightarrow\left(a-1\right)\left(a+1-2m\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=1\left(l\right)\\a=2m-1\end{matrix}\right.\)

Để pt có nghiệm \(\Leftrightarrow\left[{}\begin{matrix}2m-1\ge2\\2m-1\le-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m\ge\frac{3}{2}\\m\le-\frac{1}{2}\end{matrix}\right.\)

Lời giải:

Để hàm số xác định trên $x\in [0;2]$ thì:
\(\left\{\begin{matrix} x+2m-1\geq 0\\ 4-2m-\frac{x}{2}\geq 0\end{matrix}\right., \forall x\in [0;2]\)

\(\Leftrightarrow \left\{\begin{matrix} m\geq \frac{1-x}{2}\\ m\leq 2-\frac{x}{4}\end{matrix}\right., \forall x\in [0;2]\)

\(\Leftrightarrow \left\{\begin{matrix} m\geq \frac{1-0}{2}\\ m\leq 2-\frac{2}{4}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} m\geq \frac{1}{2}\\ m\leq \frac{3}{2}\end{matrix}\right.\)

\(\Leftrightarrow m\in [\frac{1}{2}; \frac{3}{2}]\)

Lời giải:

$2m^2+4m+4=2(m^2+2m+1)+2=2(m+1)^2+2\geq 2$ với mọi $m\in\mathbb{R}$

$\Rightarrow \sqrt{2m^2+4m+4}\geq \sqrt{2}$

$\Rightarrow A=\frac{1}{\sqrt{2m^2+4m+4}}\leq \frac{1}{\sqrt{2}}$

Vậy GTLN của $A=\frac{1}{\sqrt{2}}$ khi $m+1=0\Leftrightarrow m=-1$

Bạn hỏi hay trả lời vậy?