Initially 211, try to find the range of real number $m$ that satisfies the existence of two real numbers $a, b$, such that $a \leqslant 0, b \neq 0$, and
$$
\frac{a-b+\sqrt{a b}}{m a+3 b+\sqrt{a b}}=\frac{1}{m} \text {. }
$$
Solution: First, it must be that $m \neq 0$.
Below, we discuss the cases for $a$.
(1) If $a=0$, then $\frac{-b}{3 b}=\frac{1}{m}(m \neq 0)$.

Since $b \neq 0$, we have $-\frac{1}{3}=\frac{1}{m}$, which means $m=-3$.
(2) If $a \neq 0$, then from $a \leqslant 0$ we get $a0$.
Thus, the given equation becomes $\frac{a-t a-a \sqrt{t}}{m a+3 t a-a \sqrt{t}}=\frac{1}{m}$, that is,
$$
\begin{array}{l}
\frac{1-t-\sqrt{t}}{m+3 t-\sqrt{t}}=\frac{1}{m} \text {. } \\
\text { Hence }\left\{\begin{array}{l}
m(1-t-\sqrt{t})=m+3 t-\sqrt{t}(m \neq 0), \\
1-t-\sqrt{t} \neq 0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
t(m+3)=\sqrt{t}(1-m), \\
t+\sqrt{t}-1 \neq 0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array} { l } 
{ ( m + 3 ) \sqrt { t } = 1 - m , } \\
{ t + \sqrt { t } - 1 \neq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\sqrt{t}=\frac{1-m}{m+3}, \\
\sqrt{t} \neq \frac{-1 \pm \sqrt{5}}{2}
\end{array}\right.\right. \\
\Leftrightarrow\left\{\begin{array} { l } 
{ \sqrt { t } = \frac { 1 - m } { m + 3 } , } \\
{ \sqrt { t } \neq \frac { - 1 + \sqrt { 5 } } { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{1-m}{m+3}>0, \\
\frac{1-m}{m+3} \neq \frac{-1+\sqrt{5}}{2}
\end{array}\right.\right. \\
\Leftrightarrow\left\{\begin{array}{l}
(m+3)(1-m)>0, \\
1-m \neq \frac{-1+\sqrt{5}}{2}(m+3)
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array} { l } 
{ ( m + 3 ) ( m - 1 ) < 0 , } \\
{ \frac { 1 + \sqrt { 5 } } { 2 } m \neq \frac { 5 - 3 \sqrt { 5 } } { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
-3<m<1, \\
m \neq \frac{5-3 \sqrt{5}}{\sqrt{5}+1}
\end{array}\right.\right. \\
\Leftrightarrow\left\{\begin{array}{l}
-3<m<1, \\
m \neq-5+2 \sqrt{5}
\end{array}\right. \\
\Leftrightarrow-3<m<-5+2 \sqrt{5} \text { or }-5+2 \sqrt{5}<m<1(m \neq 0) \text {. } \\
\end{array}
$$

In summary, the range of values for $m$ is $-3 \leqslant m<-5+2 \sqrt{5}$ or $-5+2 \sqrt{5}<m<0$ or $0<m<1$.
(Wu Weizhao, School of Mathematics and Information Science, Guangzhou University, 510006).