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I wonder if $$\left\langle x\left|AB^{-1}\right |x\right\rangle = \frac{\left}{\left}$$ in general. If not, under what condition are they equal?

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    How do you define $\frac{A}{B}$ (assuming $B$ is invertible)? – 2017-02-11
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    $\frac AB$ is not unique: $AB^{-1}$ may not equal $B^{-1}A$. – 2017-02-11
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    @Clement C I would say $AB^{-1}$. – 2017-02-11
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    Then, why not $B^{-1}A$? That seems quite arbitrary for the LHS, and the RHS does not entail such choice to be made... – 2017-02-11
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    @Clement C i would say maybe limited to the case where $AB^{-1} = B^{-1}A$ and wonder if the equality holds in this case. – 2017-02-11
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    So, first restriction: $A,B^{-1}$ commute. – 2017-02-11
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    @Clement C yes.... – 2017-02-11
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    You would also need $|x \rangle$ to be a unit vector – 2017-02-12
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    @Omnomnomnom thank you for the answer. But in my case the norm of $\left|x\right>$ need not be one. I see that your answer does not assume this also. – 2017-02-13
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    Multiply $x$ by $r$. The left hand side is multiplied by $r^2$, the right hand side is multiplied by $r^0$. This strongly suggests the formula is unreasonable. It would be better to divide the l.h.s. by $\langle x|x\rangle$. – 2017-02-13

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They are not equal. They will be equal, however, if $A$ and $B$ commute and $x$ is a unit vector.

Here's a fact you may be interested in: if $B$ is positive semidefinite, $$ \frac{\left}{\left} = \frac{\left}{\left} = \frac{\left}{\left} = \frac{\left}{\left} $$