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mk chắc chắn 100% là 99m<9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
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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 \(\Leftrightarrow\left[{}\begin{matrix}2m-1\ge2\\2m-1\le-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m\ge\frac{3}{2}\\m\le-\frac{1}{2}\end{matrix}\right.\)
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Lời giải:
Để hàm số xác định trên $x\in [0;2]$ thì:
\(\left\{\begin{matrix}
x+2m-1\geq 0\\
4-2m-\frac{x}{2}\geq 0\end{matrix}\right., \forall x\in [0;2]\)
\(\Leftrightarrow \left\{\begin{matrix} m\geq \frac{1-x}{2}\\ m\leq 2-\frac{x}{4}\end{matrix}\right., \forall x\in [0;2]\)
\(\Leftrightarrow \left\{\begin{matrix} m\geq \frac{1-0}{2}\\ m\leq 2-\frac{2}{4}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} m\geq \frac{1}{2}\\ m\leq \frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow m\in [\frac{1}{2}; \frac{3}{2}]\)
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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 $m+1=0\Leftrightarrow m=-1$