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Your input: perform the partial fraction decomposition of $$$\frac{1}{\left(x - 2\right) \left(x - 1\right)^{2}}$$$

The form of the partial fraction decomposition is

$$\frac{1}{\left(x - 2\right) \left(x - 1\right)^{2}}=\frac{A}{x - 1}+\frac{B}{\left(x - 1\right)^{2}}+\frac{C}{x - 2}$$

Write the right-hand side as a single fraction:

$$\frac{1}{\left(x - 2\right) \left(x - 1\right)^{2}}=\frac{\left(x - 2\right) \left(x - 1\right) A + \left(x - 2\right) B + \left(x - 1\right)^{2} C}{\left(x - 2\right) \left(x - 1\right)^{2}}$$

The denominators are equal, so we require the equality of the numerators:

$$1=\left(x - 2\right) \left(x - 1\right) A + \left(x - 2\right) B + \left(x - 1\right)^{2} C$$

Expand the right-hand side:

$$1=x^{2} A + x^{2} C - 3 x A + x B - 2 x C + 2 A - 2 B + C$$

Collect up the like terms:

$$1=x^{2} \left(A + C\right) + x \left(- 3 A + B - 2 C\right) + 2 A - 2 B + C$$

The coefficients near the like terms should be equal, so the following system is obtained:

$$\begin{cases} A + C = 0\\- 3 A + B - 2 C = 0\\2 A - 2 B + C = 1 \end{cases}$$

Solving it (for steps, see system of equations calculator), we get that $$$A=-1$$$, $$$B=-1$$$, $$$C=1$$$

Therefore,

$$\frac{1}{\left(x - 2\right) \left(x - 1\right)^{2}}=\frac{-1}{x - 1}+\frac{-1}{\left(x - 1\right)^{2}}+\frac{1}{x - 2}$$

Answer: $$$\frac{1}{\left(x - 2\right) \left(x - 1\right)^{2}}=\frac{-1}{x - 1}+\frac{-1}{\left(x - 1\right)^{2}}+\frac{1}{x - 2}$$$