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\(1.x^2+\dfrac{1}{x^2}-2m\left(x+\dfrac{1}{x}\right)+1+2m=0\left(1\right)\)\(đặt:x^2+\dfrac{1}{x^2}=t\)
\(x>0\Rightarrow t\ge2\sqrt{x^2.\dfrac{1}{x^2}}=2\)
\(x< 0\Rightarrow-t=-x^2+\dfrac{1}{\left(-x^2\right)}\ge2\Rightarrow t\le-2\)
\(\Rightarrow t\in(-\infty;-2]\cup[2;+\infty)\left(2\right)\)
\(\Rightarrow\left(1\right)\Leftrightarrow t^2-2mt+2m-1=0\)
\(\Leftrightarrow\left(t-1\right)\left(t-2m+1\right)=0\Leftrightarrow\left[{}\begin{matrix}t=1\notin\left(2\right)\\t=2m-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2m-1\le-2\\2m-1\ge2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}m\le-\dfrac{1}{2}\\m\ge\dfrac{3}{4}\end{matrix}\right.\)
\(2.\) \(f^2\left(\left|x\right|\right)+\left(m-2\right)f\left(\left|x\right|\right)+m-3=0\left(1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}f\left(\left|x\right|\right)=-1\\f\left(\left|x\right|\right)=3-m\end{matrix}\right.\)
\(dựa\) \(vào\) \(đồ\) \(thị\) \(f\left(\left|x\right|\right)\) \(\Rightarrow f\left(\left|x\right|\right)=-1\) \(có\) \(2nghiem\) \(pb\)
\(\left(1\right)có\) \(6\) \(ngo\) \(pb\Leftrightarrow\left\{{}\begin{matrix}-1< 3-m< 3\\3-m\ne-1\\\end{matrix}\right.\)\(\Leftrightarrow0< m< 4\)
\(\Rightarrow m=\left\{1;2;3\right\}\)
Đặt \(f\left(1\right)=d\)
\(f\left(n+1\right)=af^2\left(n\right)+bf\left(n\right)+\dfrac{b^2}{4a}-\dfrac{b}{2a}\)
\(\Leftrightarrow f\left(n+1\right)+\dfrac{b}{2a}=a\left[f\left(n\right)+\dfrac{b}{2a}\right]^2\)
Đặt \(f\left(n\right)+\dfrac{b}{2a}=g\left(n\right)\Rightarrow\left\{{}\begin{matrix}g\left(1\right)=d+\dfrac{b}{2a}\\g\left(n+1\right)=a.g^2\left(n\right)\end{matrix}\right.\)
\(\Rightarrow g\left(n\right)=a.g^2\left(n-1\right)=a\left[a.g^2\left(n-2\right)\right]^2=a^{2^2-1}.g^{2^2}\left(n-2\right)=...=a^{2^{n-1}-1}.\left[g\left(1\right)\right]^{2^{n-1}}\)
\(\Rightarrow g\left(n\right)=a^{2^{n-1}-1}.\left(d+\dfrac{b}{2a}\right)^{2^{n-1}}\)
\(\Rightarrow f\left(n\right)=a^{2^{n-1}-1}.\left(d+\dfrac{b}{2a}\right)^{2^{n-1}}-\dfrac{b}{2a}\) (1)
Sau đó kiểm tra lại công thức (1) bằng quy nạp là được
bổ sung đề
với f không giảm
tính f\(\left(\frac{1}{n}\right)\) với n∈\(\left\{1;2;3;....;20\right\}\)