Answer
(a) Simplified form: $-1$; excluded value: $x \ne 2.5$
(b) Simplified form: $-y-4$; excluded value: $y \ne 4$
(c) Simplified form: $- \frac{3}{{2d + 1}}$; excluded values: $d \ne \frac{1}{3}$, $d \ne -\frac{1}{2}$
(d) Simplified form: $- \frac{3}{{2z + 2}}$; excluded values: $z \ne\pm 1$
Work Step by Step
(a)
Factor out $-1$ from the denominator and then cancel the common factor $2x-5$.
$$\begin{aligned}\frac{2x-5}{5-2x}&=\frac{2x-5}{-1(2x-5)}\\&=-1\end{aligned}$$
Set the denominator to zero.
$$\begin{aligned}5 - 2x &= 0\\5 &= 2x\\x &= 2.5\end{aligned}$$
Therefore, the simplified form is $-1$ with $x \ne 2.5$.
(b)
Factor the numerator using the difference of squares formula and factor out $-1$ from the denominator, cancel the common factor $y-4$ and simplify.
$$\begin{aligned}\frac{{{y^2} - 16}}{{4 - y}} &= \frac{{\left( {y - 4} \right)\left( {y + 4} \right)}}{{ - \left( {y - 4} \right)}}\\ &= \frac{{y + 4}}{{ - 1}}\\ &= - \left( {y + 4} \right)\\ &= - y - 4\end{aligned}$$
Set the denominator to zero.
$$\begin{aligned}4 - y &= 0\\y &= 4\end{aligned}$$
Therefore, the simplified form is $-y-4$ with $y \ne 4$.
(c)
Factor out $-3$ from the numerator and factorize the denominator.
$$\begin{aligned}\frac{{3 - 9d}}{{6{d^2} + d - 1}} &= \frac{{ - 3\left( {3d - 1} \right)}}{{6{d^2} + 3d - 2d - 1}}\\ &= \frac{{ - 3\left( {3d - 1} \right)}}{{3d\left( {2d + 1} \right) - 1\left( {2d + 1} \right)}}\\ &= \frac{{ - 3\left( {3d - 1} \right)}}{{\left( {3d - 1} \right)\left( {2d + 1} \right)}}\\ &= - \frac{3}{{2d + 1}}\end{aligned}$$
Set the denominator to zero.
$$\begin{aligned}6{d^2} + d - 1 &= 0\\6{d^2} + 3d - 2d - 1 &= 0\\3d\left( {2d + 1} \right) - 1\left( {2d + 1} \right) &= 0\\\left( {3d - 1} \right)\left( {2d + 1} \right) &= 0\\d &= - \frac{1}{2},\frac{1}{3}\end{aligned}$$
Therefore, the simplified form is $- \frac{3}{{2d + 1}}$ with $d \ne \frac{1}{3}$ and $d \ne -\frac{1}{2}$.
(d)
Factor out $-1$ from the numerator and factorize the denominator using the difference of squares formula.
$$\begin{aligned}\frac{{3 - 3z}}{{2{z^2} - 2}} &= \frac{{ - 3\left( {z - 1} \right)}}{{2\left( {{z^2} - 1} \right)}}\\ &= - \frac{{3\left( {z - 1} \right)}}{{2\left( {z - 1} \right)\left( {z + 1} \right)}}\\ &= - \frac{3}{{2\left( {z + 1} \right)}}\\ &= - \frac{3}{{2z + 2}}\end{aligned}$$
Set the denominator to zero.
$$\begin{aligned}2{z^2} - 2 &= 0\\2\left( {{z^2} - 1} \right) &= 0\\2\left( {z - 1} \right)\left( {z + 1} \right) &= 0\\z &= \pm 1\end{aligned}$$
Therefore, the simplified form is $- \frac{3}{{2z + 2}}$ with $z \ne\pm 1$.
