17. (15 points) Given the function
$$
f(x)=\frac{1}{2} m x^{2}-2 x+1+\ln (x+1)(m \geqslant 1) \text {. }
$$
(1) If the curve $C: y=f(x)$ has a tangent line $l$ at point $P(0,1)$ that intersects $C$ at only one point, find the value of $m$;
(2) Prove that the function $f(x)$ has a decreasing interval $[a, b]$, and find the range of the length $t=b-a$ of the decreasing interval.
17. (1) Note that the domain of the function $f(x)$ is $(-1,+\infty)$,
$$
f^{\prime}(x)=m x-2+\frac{1}{x+1}, f^{\prime}(0)=-1 .
$$

Therefore, the slope of the tangent line $l$ at the point of tangency $P(0,1)$ is -1.
Thus, the equation of the tangent line is $y=-x+1$.
Since the tangent line $l$ intersects the curve $C$ at only one point, the equation
$$
\frac{1}{2} m x^{2}-x+\ln (x+1)=0
$$

has exactly one real solution.
Clearly, $x=0$ is a solution to the equation.
Let $g(x)=\frac{1}{2} m x^{2}-x+\ln (x+1)$. Then
$$
g^{\prime}(x)=m x-1+\frac{1}{x+1}=\frac{m x\left[x-\left(\frac{1}{m}-1\right)\right]}{x+1} \text {. }
$$

When $m=1$, $g^{\prime}(x)=\frac{x^{2}}{x+1} \geqslant 0$ (with equality only when $x=0$), so $g(x)$ is monotonically increasing on $(-1,+\infty)$, meaning $x=0$ is the only real solution to the equation.
When $m>1$, from
$$
g^{\prime}(x)=\frac{m x\left[x-\left(\frac{1}{m}-1\right)\right]}{x+1}=0 \text {, }
$$

we get $x_{1}=0, x_{2}=\frac{1}{m}-1 \in(-1,0)$.
On the interval $\left(-1, x_{2}\right)$, $g^{\prime}(x)>0$, and on the interval $\left(x_{2}, 0\right)$, $g^{\prime}(x)g(0)=0$.

As $x \rightarrow-1$, $g(x) \rightarrow-\infty$, so $g(x)=0$ has another solution in $\left(-1, x_{2}\right)$, which is a contradiction.
In summary, we have $m=1$.
(2) Note that
$$
\begin{array}{l}
f^{\prime}(x)=\frac{m x^{2}+(m-2) x-1}{x+1}(x>-1) . \\
\text { Hence } f^{\prime}(x)0$, and the axis of symmetry is
$$
\begin{array}{l}
x=-\frac{1}{2}+\frac{1}{m}>-1, \\
h(-1)=m-(m-2)-1=1>0,
\end{array}
$$

Therefore, the equation $h(x)=0$ has two distinct real roots $x_{1} 、 x_{2}$ in $(-1,+\infty)$, meaning the solution set of equation (1) is $\left(x_{1} 、 x_{2}\right)$.
Thus, the interval of monotonic decrease for the function $f(x)$ is $\left[x_{1}, x_{2}\right]$.
Then $t=x_{2}-x_{1}=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}$
$$
=\sqrt{\frac{\Delta}{m^{2}}}=\sqrt{1+\frac{4}{m^{2}}} \text {. }
$$

Since $m \geqslant 1$, we have $1<\sqrt{1+\frac{4}{m^{2}}} \leqslant \sqrt{5}$.
Therefore, the range of the length $t$ of the interval of monotonic decrease for the function $y=f(x)$ is $(1, \sqrt{5}]$.