The problem 23 asks to simplify the expression $\frac{a}{b} - \frac{b}{a} - \frac{c}{b}$. The problem 24 describes an equilateral triangle $XYZ$ with side length 6 cm. $T$ is the midpoint of $XY$. We need to find $\tan(\angle ZXT)$.

1. Problem Description

The problem 23 asks to simplify the expression $\frac{a}{b} - \frac{b}{a} - \frac{c}{b}$.

The problem 24 describes an equilateral triangle $XYZ$ with side length 6 cm. $T$ is the midpoint of $XY$. We need to find $\tan(\angle ZXT)$.

2. Solution Steps

Problem 23:

We have $\frac{a}{b} - \frac{b}{a} - \frac{c}{b}$. To combine the fractions, we need a common denominator. The common denominator is $ab$.

$\frac{a}{b} - \frac{b}{a} - \frac{c}{b} = \frac{a^2}{ab} - \frac{b^2}{ab} - \frac{ac}{ab} = \frac{a^2 - b^2 - ac}{ab}$.

Problem 24:

Since $XYZ$ is an equilateral triangle, all angles are $60^\circ$. Thus, $\angle ZXY = 60^\circ$. Also, $XT = TY = \frac{6}{2} = 3$ cm.

Since $T$ is the midpoint of $XY$, $ZT$ is the median to the side $XY$. Also, in an equilateral triangle, the median is also an altitude. Therefore, $ZT \perp XY$, and $\angle ZTX = 90^\circ$.

Consider the right triangle $ZTX$. We want to find $\tan(\angle ZXT)$.

$\tan(\angle ZXT) = \frac{ZT}{XT}$.

We need to find the length of $ZT$. Since $XYZ$ is an equilateral triangle with side length 6, we can find the length of the altitude $ZT$. Consider triangle $ZTX$.

$ZX^2 = ZT^2 + XT^2$

$6^2 = ZT^2 + 3^2$

$36 = ZT^2 + 9$

$ZT^2 = 36 - 9 = 27$

$ZT = \sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3}$

Now we can find $\tan(\angle ZXT) = \frac{ZT}{XT} = \frac{3\sqrt{3}}{3} = \sqrt{3}$.

3. Final Answer

Problem 23: D. $\frac{a^2 - b^2 - ac}{ab}$

Problem 24: C. $\sqrt{3}$