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

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

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

$$1=\left(x - 2\right) A + \left(x - 1\right) B$$

Expand the right-hand side:

$$1=x A + x B - 2 A - B$$

Collect up the like terms:

$$1=x \left(A + B\right) - 2 A - B$$

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

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

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

Therefore,

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

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