Step 1: Simplify the numerator of the fraction. $$\frac{y^5}{y^4} = y^{5-4} = y$$ Step 2: Simplify the denominator of the fraction. $$y^4 - y^4 = 0$$ Step 3: Substitute the simplified numerator and denominator back into the expression. The expression becomes $2 - \left(\frac{y}{0}\right)$. Step 4: Evaluate the expression. Division by zero is undefined. Therefore, the entire expression is undefined. $$\boxed{\text{Undefined}}$$ Step 1: Set up the partial fraction decomposition. $$\frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$$ Step 2: Multiply both sides by the common denominator $(x-1)(x+2)$. $$1 = A(x+2) + B(x-1)$$ Step 3: Solve for $A$ and $B$ using the cover-up method or by equating coefficients. To find $A$, set $x=1$: $$1 = A(1+2) + B(1-1)$$ $$1 = 3A + 0$$ $$A = \frac{1}{3}$$ To find $B$, set $x=-2$: $$1 = A(-2+2) + B(-2-1)$$ $$1 = 0 + B(-3)$$ $$B = -\frac{1}{3}$$ Step 4: Substitute the values of $A$ and $B$ back into the partial fraction decomposition. $$\frac{1}{(x-1)(x+2)} = \frac{\frac{1}{3}}{x-1} + \frac{-\frac{1}{3}}{x+2}$$ $$\frac{1}{(x-1)(x+2)} = \frac{1}{3(x-1)} - \frac{1}{3(x+2)}$$ $$\boxed{\frac{1}{3(x-1)} - \frac{1}{3(x+2)}}$$ Step 1: Set up the partial fraction decomposition. For a repeated linear factor $(x-2)^2$, the decomposition includes terms $\frac{A}{x-2}$ and $\frac{B}{(x-2)^2}$. $$\frac{3}{(x-2)^2(x+1)} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{x+1}$$ Step 2: Multiply both sides by the common denominator $(x-2)^2(x+1)$. $$3 = A(x-2)(x+1) + B(x+1) + C(x-2)^2$$ Step 3: Solve for $A$, $B$, and $C$ using strategic values of $x$. To find $B$, set $x=2$: $$3 = A(2-2)(2+1) + B(2+1) + C(2-2)^2$$ $$3 = A(0) + B(3) + C(0)$$ $$3 = 3B \Rightarrow B = 1$$ To find $C$, set $x=-1$: $$3 = A(-1-2)(-1+1) + B(-1+1) + C(-1-2)^2$$ $$3 = A(0) + B(0) + C(-3)^2$$ $$3 = 9C \Rightarrow C = \frac{3}{9} = \frac{1}{3}$$ To find $A$, choose a convenient value for $x$, such as $x=0$, and substitute the known values of $B$ and $C$: $$3 = A(0-2)(0+1) + B(0+1) + C(0-2)^2$$ $$3 = A(-2)(1) + B(1) + C(-2)^2$$ $$3 = -2A + B + 4C$$ Substitute $B=1$ and $C=\frac{1}{3}$: $$3 = -2A + 1 + 4\left(\frac{1}{3}\right)$$ $$3 = -2A + 1 + \frac{4}{3}$$ $$3 - 1 - \frac{4}{3} = -2A$$ $$2 - \frac{4}{3} = -2A$$ $$\frac{6}{3} - \frac{4}{3} = -2A$$ $$\frac{2}{3} = -2A$$ $$A = \frac{2}{3} \times \left(-\frac{1}{2}\right) = -\frac{1}{3}$$ Step 4: Substitute the values of $A$, $B$, and $C$ back into the partial fraction decomposition. $$\frac{3}{(x-2)^2(x+1)} = \frac{-\frac{1}{3}}{x-2} + \frac{1}{(x-2)^2} + \frac{\frac{1}{3}}{x+1}$$ $$\frac{3}{(x-2)^2(x+1)} = -\frac{1}{3(x-2)} + \frac{1}{(x-2)^2} + \frac{1}{3(x+1)}$$ $$\boxed{-\frac{1}{3(x-2)} + \frac{1}{(x-2)^2} + \frac{1}{3(x+1)}}$$