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