Cho hàm số $y=f\left( x \right)$ liên tục trên $\mathbb{R}$ và có...

Ta có
Khi đó: $f\left( f\left( x \right)-1 \right)=0\Leftrightarrow \left[ \begin{matrix}
f\left( x \right)-1={{x}_{1}}\in \left( -2 ; -1 \right) \\
f\left( x \right)-1={{x}_{2}}\in \left( -1 ; 0 \right) \\
f\left( x \right)-1={{x}_{3}}\in \left( 1; 2 \right) \\
\end{matrix} \right.$$\Leftrightarrow \left[ \begin{matrix}
f\left( x \right)=1+{{x}_{1}}\in \left( -1; 0 \right) \\
f\left( x \right)=1+{{x}_{2}}\in \left( 0 ; 1 \right) \\
f\left( x \right)=1+{{x}_{3}}\in \left( 2; 3 \right) \\
\end{matrix} \right.f\left( x \right)=1+{{x}_{1}}\in \left( -1; 0 \right)f\left( x \right)=1+{{x}_{2}}\in \left( 0 ; 1 \right)f\left( x \right)=1+{{x}_{3}}\in \left( 2; 3 \right)f\left( f\left( x \right)-1 \right)=07$ nghiệm.