Tìm Hàm số \(f\left(x\right)\), biết :
\(\left(x-1\right)f\left(x\right)+f\left(\frac{1}{x}\right)=\frac{1}{x-1}\) (\(x\ne0\); \(x\ne1\))
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4 tháng 9 2019
\(f\left(x\right)=\frac{x^2+2x+1-x^2}{x^2\left(x+1\right)^2}=\frac{\left(x+1\right)^2-x^2}{x^2\left(x+1\right)^2}=\frac{1}{x^2}-\frac{1}{\left(x+1\right)^2}\)
\(\Rightarrow f\left(1\right)+f\left(2\right)+....+f\left(x\right)=1-\frac{1}{2^2}+\frac{1}{2^2}-....-\frac{1}{\left(x+1\right)^2}\)
\(\Rightarrow\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-19+x=\frac{x\left(x+2\right)}{\left(x+1\right)^2}\)
\(\Leftrightarrow\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-19+x=\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-20+\left(x+1\right)=\frac{x\left(x+2\right)}{\left(x+1\right)^2}\)
Dat:\(x+1=a\Rightarrow\frac{\left(2y+1\right)a^3-20a^2-1}{a^2}=\frac{a^2-1}{a^2}\Leftrightarrow\left(2y+1\right)a^3-20a^2-1=a^2-1\)
\(\Leftrightarrow\left(2y+1\right)a^3-20a^2=a^2\Leftrightarrow\left(2ay+a\right)-20=1\left(coi:x=-1cophailanghiemko\right)\)
\(\Leftrightarrow2ay+a=21\Leftrightarrow a\left(2y+1\right)=21\Leftrightarrow\left(x+1\right)\left(2y+1\right)=21\)
15 tháng 10 2020
Hướng dẫn:
Đặt: \(\frac{1}{x}=t\)( t khác 0; 1)
=> \(f\left(1-t\right)+2f\left(t\right)=\frac{1}{t}\)=> \(2f\left(1-t\right)+4f\left(t\right)=\frac{2}{t}\)(1)
Đặt: \(\frac{1}{x}=1-t\)
=> \(f\left(t\right)+2f\left(1-t\right)=\frac{1}{1-t}\)(2)
Lấy (1) - (2) => \(f\left(t\right)=\frac{1}{3}\left(\frac{2}{t}-\frac{1}{1-t}\right)\)
Vậy \(f\left(x\right)=\frac{1}{3}\left(\frac{2}{x}-\frac{1}{1-x}\right)\)
P/s: Chú ý điều kiện
29 tháng 1 2022
\(1.x^2+\dfrac{1}{x^2}-2m\left(x+\dfrac{1}{x}\right)+1+2m=0\left(1\right)\)\(đặt:x^2+\dfrac{1}{x^2}=t\)
\(x>0\Rightarrow t\ge2\sqrt{x^2.\dfrac{1}{x^2}}=2\)
\(x< 0\Rightarrow-t=-x^2+\dfrac{1}{\left(-x^2\right)}\ge2\Rightarrow t\le-2\)
\(\Rightarrow t\in(-\infty;-2]\cup[2;+\infty)\left(2\right)\)
\(\Rightarrow\left(1\right)\Leftrightarrow t^2-2mt+2m-1=0\)
\(\Leftrightarrow\left(t-1\right)\left(t-2m+1\right)=0\Leftrightarrow\left[{}\begin{matrix}t=1\notin\left(2\right)\\t=2m-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2m-1\le-2\\2m-1\ge2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}m\le-\dfrac{1}{2}\\m\ge\dfrac{3}{4}\end{matrix}\right.\)
\(2.\) \(f^2\left(\left|x\right|\right)+\left(m-2\right)f\left(\left|x\right|\right)+m-3=0\left(1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}f\left(\left|x\right|\right)=-1\\f\left(\left|x\right|\right)=3-m\end{matrix}\right.\)
\(dựa\) \(vào\) \(đồ\) \(thị\) \(f\left(\left|x\right|\right)\) \(\Rightarrow f\left(\left|x\right|\right)=-1\) \(có\) \(2nghiem\) \(pb\)
\(\left(1\right)có\) \(6\) \(ngo\) \(pb\Leftrightarrow\left\{{}\begin{matrix}-1< 3-m< 3\\3-m\ne-1\\\end{matrix}\right.\)\(\Leftrightarrow0< m< 4\)
\(\Rightarrow m=\left\{1;2;3\right\}\)
......................?
mik ko biết
mong bn thông cảm
nha ................
Ta có : \(\left(x-1\right)f\left(x\right)+f\left(\frac{1}{x}\right)=\frac{1}{x-1}\) (1)
Thay x bởi \(\frac{1}{x}\)thì đẳng thức thành :
\(\left(\frac{1}{x}-1\right)f\left(\frac{1}{x}\right)+f\left(x\right)=\frac{1}{\frac{1}{x}-1}\)
Hay : \(\frac{1-x}{x}f\left(\frac{1}{x}\right)+f\left(x\right)=\frac{x}{1-x}\) (2)
Nhân \(\frac{1-x}{x}\)vào hai vế của (1), ta được :
\(\frac{-x^2+2x-1}{x}f\left(x\right)+\frac{1-x}{x}f\left(\frac{1}{x}\right)=-\frac{1}{x}\) (3)
Lấy (2) trừ đì (3) theo từng vế, ta được :
\(\left[1-\frac{-x^2+2x-1}{x}\right]f\left(x\right)=\frac{x}{1-x}+\frac{1}{x}\)
Suy ra : \(f\left(x\right)=\frac{1}{1-x}\)