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a) Với x1 = x2 = 1
\(\Rightarrow f\left(1\right)=f\left(1.1\right)\)
\(\Rightarrow f\left(1\right)=f\left(1\right).f\left(1\right)\)
\(\Rightarrow f\left(1\right).f\left(1\right)-f\left(1\right)=0\)
\(\Rightarrow f\left(1\right).\left[f\left(1\right)-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}f\left(1\right)=0\\f\left(1\right)-1=0\end{cases}}\)
Mà \(f\left(x\right)\ne0\) ( với mọi \(x\in R\) \(;\) \(x\ne0\) )
\(\Rightarrow f\left(1\right)\ne0\)
\(\Rightarrow f\left(1\right)-1=0\)
\(\Rightarrow f\left(1\right)=1\)
b) Ta có : \(f\left(\frac{1}{x}\right).f\left(x\right)=f\left(\frac{1}{x}.x\right)\)
\(\Rightarrow f\left(\frac{1}{x}\right).f\left(x\right)=f\left(1\right)=1\)
\(\Rightarrow f\left(\frac{1}{x}\right).f\left(x\right)=1\)
\(\Rightarrow f\left(\frac{1}{x}\right)=\frac{1}{f\left(x\right)}\)
\(\Rightarrow f\left(x^{-1}\right)=\left[f\left(x\right)\right]^{-1}\)
a)Với x1 = x2 = 1
\( \implies\) \(f\left(1\right)=f\left(1.1\right)\)
\( \implies\) \(f\left(1\right)=f\left(1\right).f\left(1\right)\)
\( \implies\)\(f\left(1\right).f\left(1\right)-f\left(1\right)=0\)
\( \implies\) \(f\left(1\right).\left[f\left(1\right)-1\right]=0\)
\( \implies\) \(\orbr{\begin{cases}f\left(1\right)=0\\f\left(1\right)-1=0\end{cases}}\)
Mà \(f\left(x\right)\) khác \(0\) ( với mọi \(x\) \(\in\) \(R\) ; \(x\) khác \(0\) )
\( \implies\) \(f\left(1\right)\) khác \(0\)
\( \implies\) \(f\left(1\right)-1=0\)
\( \implies\) \(f\left(1\right)=1\)
b)Ta có : \(f\left(\frac{1}{x}\right).f\left(x\right)=f\left(\frac{1}{x}.x\right)\)
\( \implies\) \(f\left(\frac{1}{x}\right).f\left(x\right)=f\left(1\right)=1\)
\( \implies\) \(f\left(\frac{1}{x}\right).f\left(x\right)=1\)
\( \implies\) \(f\left(\frac{1}{x}\right)=\frac{1}{f\left(x\right)}\)
\( \implies\) \(f\left(x^{-1}\right)=\left[f\left(x\right)\right]^{-1}\)
a)Cho y=f(x)=-5x
C/m f(\(x_1\)+\(4x_2\))=f(\(x_1\))+4f(x2)
b)Cho f(x)+3f(\(\frac{1}{2}\))=x2.Tính f(2)
Xin lỗi, (1) xảy ra khi x,(x-8) cùng dấu.
Ta có:(1)<=>\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x-8>0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-8< 0\end{matrix}\right.\end{matrix}\right.\)<=>\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x>8\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< 8\end{matrix}\right.\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x>8\\x< 0\end{matrix}\right.\)
Vậy x>8 hoặc x<0.
1) Theo đề bài: x2-8x+9>9
<=>x2-8x>0
<=>x(x-8)>0(1)
(1) xảy ra khi x;(x-8) trái dấu.
Mà x>x-8 với mọi x nên:
(1)<=>\(\left\{{}\begin{matrix}x>0\\x-8< 0\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}x>0\\x< 8\end{matrix}\right.\)<=>0<x<8
Vậy 0<x<8.
Phần này khó chú ý nè bạn
Giải
Ta có f(x1+x2) = f(x1) + f(x2)
nên f(7) = f(3)+f(4)= f(2)+f(1) + f(2)+f(2) = f(1)+f(1)+f(1)+f(1)+f(1)+f(1)+f(1)=7
\(f\left(\dfrac{1}{7}\right)=\dfrac{1}{49}.f\left(7\right)=\dfrac{1}{49}.7=\dfrac{1}{7}\)
Ta có :\(f\left(\dfrac{5}{7}\right)=f\left(\dfrac{2}{7}\right)+f\left(\dfrac{3}{7}\right)=f\left(\dfrac{1}{7}\right)+f\left(\dfrac{1}{7}\right)+f\left(\dfrac{1}{7}\right)+f\left(\dfrac{2}{7}\right)=f\left(\dfrac{1}{7}\right)+f\left(\dfrac{1}{7}\right)+f\left(\dfrac{1}{7}\right)+f\left(\dfrac{1}{7}\right)+f\left(\dfrac{1}{7}\right)=\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{7}=\dfrac{5}{7}\)
chụp lại bài nhìn cho rõ hơn .