Sure! Let's solve each equation one by one:

**Equation 9: \(x^2 = x^5\)**

Divide both sides by \(x^2\) (assuming \(x \neq 0\)):

\[
1 = x^3
\]

So, \(x = 1\).

Also, check for \(x = 0\): the equation \(0^2 = 0^5\) holds true.

So, the solutions for equation 9 are \(x = 0\) and \(x = 1\).

**Equation 10: \((3y - 1)^{10} = (3y - 1)^{20}\)**

Assuming \(3y - 1 \neq 0\), divide both sides by \((3y - 1)^{10}\):

\[
1 = (3y - 1)^{10}
\]

This implies \((3y - 1) = 1\), so:

\[
3y - 1 = 1 \implies 3y = 2 \implies y = \frac{2}{3}
\]

For \(3y - 1 = 0\):

\[
3y - 1 = 0 \implies 3y = 1 \implies y = \frac{1}{3}
\]

So, the solutions for equation 10 are \(y = \frac{2}{3}\) and \(y = \frac{1}{3}\).

**Equation 11: \((x - 5)^2 = (1 - 3x)^2\)**

Taking square roots of both sides (considering both positive and negative roots):

\[
x - 5 = 1 - 3x \quad \text{or} \quad x - 5 = -(1 - 3x)
\]

First case:

\[
x - 5 = 1 - 3x \implies 4x = 6 \implies x = \frac{3}{2}
\]

Second case:

\[
x - 5 = -1 + 3x \implies -4 = 2x \implies x = -2
\]

So, the solutions for equation 11 are \(x = \frac{3}{2}\) and \(x = -2\).