Simplify the expression: \(2 - (y5)/(4)y(1)/(4) - y^-(1)/(4)\).

Here are the solutions to the problems.

Question 5:
Simplify $\left(2 - \frac{y^{\frac{5}{4}}}{y^{\frac{1}{4}} - y^{-\frac{1}{4}}}\right)$.

Step 1: Simplify the denominator of the fraction.

$$y^{\frac{1}{4}} - y^{-\frac{1}{4}} = y^{\frac{1}{4}} - \frac{1}{y^{\frac{1}{4}}}$$

Find a common denominator:

$$= \frac{(y^{\frac{1}{4}})^2 - 1}{y^{\frac{1}{4}}} = \frac{y^{\frac{1}{2}} - 1}{y^{\frac{1}{4}}}$$

Step 2: Substitute the simplified denominator back into the fraction and simplify.

$$\frac{y^{\frac{5}{4}}}{y^{\frac{1}{4}} - y^{-\frac{1}{4}}} = \frac{y^{\frac{5}{4}}}{\frac{y^{\frac{1}{2}} - 1}{y^{\frac{1}{4}}}}$$

Multiply by the reciprocal of the denominator:

$$= y^{\frac{5}{4}} \times \frac{y^{\frac{1}{4}}}{y^{\frac{1}{2}} - 1}$$

Combine the terms in the numerator using exponent rules ($a^m \times a^n = a^{m+n}$):

$$= \frac{y^{\frac{5}{4} + \frac{1}{4}}}{y^{\frac{1}{2}} - 1} = \frac{y^{\frac{6}{4}}}{y^{\frac{1}{2}} - 1} = \frac{y^{\frac{3}{2}}}{y^{\frac{1}{2}} - 1}$$

Step 3: Substitute this simplified fraction back into the original expression and combine terms.

$$2 - \frac{y^{\frac{3}{2}}}{y^{\frac{1}{2}} - 1}$$

Find a common denominator, which is $y^{\frac{1}{2}} - 1$:

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

Rearrange the terms in the numerator:

$$= \frac{-y^{\frac{3}{2}} + 2y^{\frac{1}{2}} - 2}{y^{\frac{1}{2}} - 1}$$

This can also be written by multiplying the numerator and denominator by $-1$:

$$= \frac{y^{\frac{3}{2}} - 2y^{\frac{1}{2}} + 2}{1 - y^{\frac{1}{2}}}$$

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

Question 6:
Split $\frac{1}{(x-1)(x+2)}$ into partial fractions.

Step 1: Set up the partial fraction decomposition.
For distinct linear factors, the form is:

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

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.
To find A, let $x=1$:

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

To find B, let $x=-2$:

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

Step 4: Substitute the values of A and B back into the partial fraction form.

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

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

Question 7:
Split $\frac{3}{(x-2)^2(x+1)}$ into partial fractions.

Step 1: Set up the partial fraction decomposition.
For a repeated linear factor $(x-2)^2$ and a distinct linear factor $(x+1)$, the form is:

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

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.
To find B, let $x=2$:

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

To find C, let $x=-1$:

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

To find A, substitute $B=1$, $C=\frac{1}{3}$ and choose a convenient value for $x$, for example $x=0$:

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