The problem has three parts: (i) Simplify the expression $(2xy^{-2})(4x^{-4}y)$. (ii) Differentiate $y = (3x+1)^3$ with respect to $x$. (iii) Solve the quadratic equation $5x^2 - 11x + 2 = 0$.

1. Problem Description

The problem has three parts:

(i) Simplify the expression $(2xy^{-2})(4x^{-4}y)$.

(ii) Differentiate $y = (3x+1)^3$ with respect to $x$.

(iii) Solve the quadratic equation $5x^2 - 11x + 2 = 0$.

2. Solution Steps

(i) Simplify $(2xy^{-2})(4x^{-4}y)$.

Multiply the coefficients and add the exponents of like variables.

$$(2xy^{-2})(4x^{-4}y) = (2 \cdot 4)(x^{1+(-4)})(y^{-2+1}) = 8x^{-3}y^{-1} = \frac{8}{x^3y}$$

(ii) Differentiate $y = (3x+1)^3$ with respect to $x$.

We use the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$. Let $u = 3x+1$, then $y = u^3$.

$$\frac{dy}{du} = 3u^2$$

$$\frac{du}{dx} = 3$$

Therefore,

$$\frac{dy}{dx} = 3u^2 \cdot 3 = 9u^2 = 9(3x+1)^2 = 9(9x^2 + 6x + 1) = 81x^2 + 54x + 9$$

(iii) Solve the quadratic equation $5x^2 - 11x + 2 = 0$.

We can use the quadratic formula to solve for $x$:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

In this case, $a = 5$, $b = -11$, and $c = 2$.

$$x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(5)(2)}}{2(5)} = \frac{11 \pm \sqrt{121 - 40}}{10} = \frac{11 \pm \sqrt{81}}{10} = \frac{11 \pm 9}{10}$$

Thus, we have two possible solutions:

$$x_1 = \frac{11 + 9}{10} = \frac{20}{10} = 2$$

$$x_2 = \frac{11 - 9}{10} = \frac{2}{10} = \frac{1}{5}$$

3. Final Answer

(i) $\frac{8}{x^3y}$

(ii) $81x^2 + 54x + 9$

(iii) $x=2$ or $x=\frac{1}{5}$