Question

[Nguyễn Lê Anh]

Cho hàm số y= f(x)\(\ne\)0 (\(\forall\)x\(\inℝ\)\(x\ne0\)) sao cho f(x₁x₂)= f(x₁).f(x₂). CMR

a) f(1)= 1

b) f(x^-1)= [f(x)]^-1

Answer 1

[I - Vy Nguyễn]

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 : \(f\left(\frac{1}{x}\right).f\left(x\right)=f\left(\frac{1}{x}.x\right)\)

\(\Rightarrow f\left(\frac{1}{x}\right).f\left(x\right)=f\left(1\right)=1\)

\(\Rightarrow f\left(\frac{1}{x}\right).f\left(x\right)=1\)

\(\Rightarrow f\left(\frac{1}{x}\right)=\frac{1}{f\left(x\right)}\)

\(\Rightarrow f\left(x^{-1}\right)=\left[f\left(x\right)\right]^{-1}\)