Solve number 5-7 na please

$$\frac{\frac{y^3}{x^4}}{\frac{y^4-y^3}{x^4}} = \frac{y^3}{x^4} \cdot \frac{x^4}{y^4-y^3}$$

Cancel $x^4$:

$$= \frac{y^3}{y^4-y^3}$$

Factor out $y^3$ from the denominator:

$$= \frac{y^3}{y^3(y-1)}$$

Cancel $y^3$ (assuming $y \neq 0$):

$$= \frac{1}{y-1}$$

$$= \frac{2(y-1)}{y-1} - \frac{1}{y-1}$$ $$= \frac{2y-2-1}{y-1}$$ $$= \frac{2y-3}{y-1}$$

The simplified expression is $\boxed{\frac{2y-3}{y-1}}$.

Number 6:

Set $x=1$:

$$1 = A(1+2) + B(1-1)$$ $$1 = 3A \implies A = \frac{1}{3}$$

Set $x=-2$:

$$1 = A(-2+2) + B(-2-1)$$ $$1 = -3B \implies B = -\frac{1}{3}$$

$$\frac{1}{(x-1)(x+2)} = \frac{\frac{1}{3}}{x-1} + \frac{-\frac{1}{3}}{x+2}$$ $$= \frac{1}{3(x-1)} - \frac{1}{3(x+2)}$$

The partial fraction decomposition is $\boxed{\frac{1}{3(x-1)} - \frac{1}{3(x+2)}}$.

Number 7:

Set $x=2$:

$$3 = A(2-2)(2+1) + B(2+1) + C(2-2)^2$$ $$3 = 0 + 3B + 0 \implies 3B = 3 \implies B = 1$$

Set $x=-1$:

$$3 = A(-1-2)(-1+1) + B(-1+1) + C(-1-2)^2$$ $$3 = 0 + 0 + C(-3)^2 \implies 3 = 9C \implies C = \frac{3}{9} = \frac{1}{3}$$

Set $x=0$ and substitute $B=1$, $C=\frac{1}{3}$:

$$3 = A(0-2)(0+1) + B(0+1) + C(0-2)^2$$ $$3 = -2A + B + 4C$$ $$3 = -2A + 1 + 4\left(\frac{1}{3}\right)$$ $$3 = -2A + 1 + \frac{4}{3}$$ $$3 - 1 - \frac{4}{3} = -2A$$ $$2 - \frac{4}{3} = -2A$$ $$\frac{6}{3} - \frac{4}{3} = -2A$$ $$\frac{2}{3} = -2A$$ $$A = \frac{2}{3} \cdot \left(-\frac{1}{2}\right) = -\frac{1}{3}$$

$$\frac{3}{(x-2)^2(x+1)} = \frac{-\frac{1}{3}}{x-2} + \frac{1}{(x-2)^2} + \frac{\frac{1}{3}}{x+1}$$ $$= -\frac{1}{3(x-2)} + \frac{1}{(x-2)^2} + \frac{1}{3(x+1)}$$

The partial fraction decomposition is $\boxed{-\frac{1}{3(x-2)} + \frac{1}{(x-2)^2} + \frac{1}{3(x+1)}}$.