Example 6 Real numbers $a, b$ satisfy $ab=1$, and let
$$
M=\frac{1}{1+a}+\frac{1}{1+b}, N=\frac{a}{1+a}+\frac{b}{1+b}.
$$

Then the relationship between $M$ and $N$ is ( )(to avoid ambiguity, add $a \neq-1, b \neq-1$).
(A) $M>N$
(B) $M=N$
(C) $M<N$
(D) Uncertain
(1996, National Junior High School Mathematics Competition)
The answer to this question is (B). Someone (see reference [1]) derived $M=N$ by taking special values of $a$ and $b$, which can only negate (A) and (C), but cannot negate (D). Choosing the correct answer is also "partially correct", with a logical flaw.

The author provides multiple solutions attached to the problem, aiming to demonstrate various methods for proving equality or comparing magnitudes.
Solution 1: (Transform one side to get the other side of the equation)
$$
\begin{aligned}
M & =\frac{b}{b+a b}+\frac{a}{a+a b} \text{ (transform) } \\
& =\frac{b}{1+b}+\frac{a}{1+a} \text{ (substitute } a b=1) \\
& =N .
\end{aligned}
$$

Solution 2: (Calculate specific values to compare magnitudes)
$$
\begin{aligned}
M & =\frac{1}{1+a}+\frac{a}{a+a b} \text{ (transform) } \\
& =\frac{1}{1+a}+\frac{a}{a+1} \text{ (substitute } a b=1) \\
& =1 \text{ (substituting } b=\frac{1}{a} \text{ has the same effect) } \\
N & =\frac{a}{1+a}+\frac{a b}{a+a b} \text{ (transform) } \\
& =\frac{a}{1+a}+\frac{1}{a+1} \text{ (substitute } a b=1 \text{ ) } \\
& =1 .
\end{aligned}
$$

Thus, $M=N$.
(You can also find the values of $M$ and $N$ by finding a common denominator.)
Solution 3: (Compare magnitudes by taking the difference)
$$
\begin{array}{l}
M-N=\frac{1-a}{1+a}+\frac{1-b}{1+b} \text{ (take the difference) } \\
=\frac{a b-a}{a b+a}+\frac{1-b}{1+b} \text{ (substitute } a b=1 \text{ ) } \\
=\frac{b-1}{b+1}+\frac{1-b}{1+b} \text{ (cancel out } a \text{ ) } \\
=0 .
\end{array}
$$

Thus, $M=N$.
Solution 4: From $a b=1$, we know
$$
\frac{1}{1+a}=\frac{a b}{a b+a}=\frac{b}{b+1} \text{. }
$$

Similarly, $\frac{1}{1+b}=\frac{a}{a+1}$.
Adding them up, we get $M=N$.
There are more equivalent transformation solutions. Below, we will take a different perspective to explain the mathematical reason why $M=N$, which turns out to be the sum of the two roots of the same quadratic equation in two different forms (see reference [2]).
Solution 5: When $a=b$, $a=b=1$, we have
$M=N=1$.
When $a \neq b$, consider $a$ and $b$ as the two real roots of the quadratic equation
$$
x^{2}+k x+1=0(|k|>2)
$$

Now, construct quadratic equations with $\frac{1}{1+a}$ and $\frac{1}{1+b}$ as roots, and with $\frac{a}{1+a}$ and $\frac{b}{1+b}$ as roots, respectively. Let
$$
\begin{array}{l}
y=\frac{1}{1+x} \Rightarrow x=\frac{1-y}{y}, \\
y=\frac{x}{1+x} \Rightarrow x=\frac{y}{1-y} . \\
\text{Substitute into equation (1), we get} \\
\left(\frac{1-y}{y}\right)^{2}+k\left(\frac{1-y}{y}\right)+1=0, \\
\left(\frac{y}{1-y}\right)^{2}+k\left(\frac{y}{1-y}\right)+1=0 .
\end{array}
$$

These two equations can be transformed into each other, thus, the sum of the roots is the same.
In fact, equations (2) and (3) are both
$$
(1-y)^{2}+k(1-y) y+y^{2}=0 \text{, }
$$

which is $y^{2}-y+\frac{1}{2-k}=0$.
Not only can we derive
$$
\frac{1}{1+a}+\frac{1}{1+b}=\frac{a}{1+a}+\frac{b}{1+b}=1,
$$

but also
$$
\frac{1}{1+a} \cdot \frac{1}{1+b}=\frac{a}{1+a} \cdot \frac{b}{1+b}=\frac{1}{2-k} .
$$
3.2 Origin

This problem was created based on the analysis of a common problem.

Translating the above text into English, please retain the original text's line breaks and formatting, and output the translation result directly.