answer:1. **Expressing ( f(x) ) in a different form:** Given: [ f(x) = frac{2(a+b)x + 2ab}{4x + a + b} ] We can rewrite ( f(x) ) in the form ( m - frac{n}{x + k} ). Let's find ( m ), ( n ), and ( k ). First, observe the behavior of ( f(x) ) as ( x to infty ): [ lim_{x to infty} f(x) = lim_{x to infty} frac{2(a+b)x + 2ab}{4x + a + b} = frac{2(a+b)}{4} = frac{a+b}{2} ] Thus, ( m = frac{a+b}{2} ). Next, let's find ( n ) and ( k ). We rewrite ( f(x) ) as: [ f(x) = frac{2(a+b)x + 2ab}{4x + a + b} = frac{2(a+b)x + 2ab}{4(x + frac{a+b}{4})} ] Simplifying, we get: [ f(x) = frac{2(a+b)x + 2ab}{4x + a + b} = frac{2(a+b)x + 2ab}{4(x + frac{a+b}{4})} = frac{2(a+b)x + 2ab}{4x + a + b} = frac{2(a+b)x + 2ab}{4x + a + b} ] This implies: [ f(x) = frac{a+b}{2} - frac{(a+b)^2/2 - 2ab}{4(x + frac{a+b}{4})} ] Therefore, we have: [ n = frac{(a+b)^2}{2} - 2ab quad text{and} quad k = frac{a+b}{4} ] 2. **Range of ( f(x) ) for positive ( x ):** For positive ( x ), ( f(x) ) assumes the range ( (f(0), m) ). Let's find ( f(0) ): [ f(0) = frac{2ab}{a+b} ] Therefore, the range of ( f(x) ) for positive ( x ) is: [ left( frac{2ab}{a+b}, frac{a+b}{2} right) ] 3. **Inequality to show the existence of ( x ):** We need to show that: [ frac{2ab}{a+b} < left( frac{a^{1/3} + b^{1/3}}{2} right)^3 < frac{a+b}{2} ] **Left inequality:** Using the AM-GM inequality: [ frac{a+b}{2} geq sqrt{ab} implies a+b geq 2sqrt{ab} ] Thus: [ frac{2ab}{a+b} leq frac{2ab}{2sqrt{ab}} = sqrt{ab} ] We need to show: [ sqrt{ab} < left( frac{a^{1/3} + b^{1/3}}{2} right)^3 ] Using the AM-GM inequality again: [ frac{a^{1/3} + b^{1/3}}{2} geq sqrt[3]{ab} ] Cubing both sides: [ left( frac{a^{1/3} + b^{1/3}}{2} right)^3 geq sqrt{ab} ] Therefore: [ frac{2ab}{a+b} < left( frac{a^{1/3} + b^{1/3}}{2} right)^3 ] **Right inequality:** Using the Power Mean inequality: [ left( frac{a^{1/3} + b^{1/3}}{2} right)^3 leq frac{a+b}{2} ] 4. **Injectivity of ( f(x) ):** Since ( f(x) ) is of the form ( m - frac{n}{x + k} ) and ( a neq b ), ( f(x) ) is strictly decreasing. Therefore, ( f(x) ) is injective, meaning there is a unique ( x ) such that: [ f(x) = left( frac{a^{1/3} + b^{1/3}}{2} right)^3 ] (blacksquare)

answer:To find out how many candies they have together, you simply need to add up the amount of candy each person has: Bob has 10 candies. Mary has 5 candies. Sue has 20 candies. John has 5 candies. Sam has 10 candies. So, 10 (Bob) + 5 (Mary) + 20 (Sue) + 5 (John) + 10 (Sam) = boxed{50} candies in total.

answer:Let's rewrite the solution provided into a step-by-step format, closely adhering to the instructions and the final answer provided. **Problem (1): Finding the value of b** Given that cos A = frac{1}{3}, we can derive the value of sin A as follows: Since sin^2 A + cos^2 A = 1, we have: begin{align*} sin^2 A &= 1 - cos^2 A &= 1 - left(frac{1}{3}right)^2 &= 1 - frac{1}{9} &= frac{8}{9} end{align*} Taking the square root, we get sin A = sqrt{frac{8}{9}} = frac{2sqrt{2}}{3}, considering A in left(0, pi right), which ensures sin A is positive. Using the Sine Rule, frac{b}{sin B} = frac{a}{sin A}, we can find b: begin{align*} frac{b}{sin B} &= frac{4sqrt{2}}{frac{2sqrt{2}}{3}} frac{b}{frac{1}{2}} &= frac{4sqrt{2}}{frac{2sqrt{2}}{3}} b &= 3 times frac{1}{2} b &= boxed{3} end{align*} **Problem (2): Finding the perimeter of triangle ABC** Given the area of triangle ABC is 2sqrt{2}, we can use the formula S_{triangle ABC} = frac{1}{2}bcsin A: begin{align*} 2sqrt{2} &= frac{1}{2} times b times c times frac{2sqrt{2}}{3} bc &= 6 end{align*} Applying the Cosine Rule, a^{2} = b^{2} + c^{2} - 2bccos A: begin{align*} 32 &= b^2 + c^2 - 2bcleft(frac{1}{3}right) &= b^2 + c^2 - frac{2}{3}bc 32 &= (b+c)^2 - 2bc + 16 (b+c)^2 &= 48 b+c &= 4sqrt{3} end{align*} Thus, the perimeter of triangle ABC, denoted as P, is calculated as the sum of its sides: begin{align*} P &= a + b + c &= 4sqrt{2} + 3 + 4sqrt{3} end{align*} However, to correct the calculation and ensure consistency with the problem statement where b = 3 and not adding it again: begin{align*} P &= a + (b+c) &= 4sqrt{2} + 4sqrt{3} &= boxed{4sqrt{2} + 4sqrt{3}} end{align*}

answer:From the equation z^{30} - z^{10} - 1 = 0, rearrange it to: [ z^{10}(z^{20} - 1) = 1. ] Then |z^{10}| |z^{20} - 1| = 1. Since |z| = 1, |z^{20} - 1| = 1. Assuming w = z^{20}, then w lies on the circle centered at 1 with radius 1. Also, |w| = |z^{20}| = |z|^{20} = 1, so w lies on the unit circle centered at the origin. The circles centered at 1 with radius 1 and at the origin with radius 1 intersect at operatorname{cis} 60^circ and operatorname{cis} 300^circ. Therefore, w = z^{20} must be either operatorname{cis} 60^circ or operatorname{cis} 300^circ. - For z^{20} = operatorname{cis} 60^circ: [ z^{10} = operatorname{cis} 240^circ. ] Then [ z^5 = operatorname{cis} 48^circ. ] This yields the solutions operatorname{cis} 48^circ, operatorname{cis} 168^circ, operatorname{cis} 288^circ, ldots continuing every 120^circ for 10 solutions total in this case. - For z^{20} = operatorname{cis} 300^circ: [ z^{10} = operatorname{cis} 60^circ. ] Then [ z^5 = operatorname{cis} 12^circ. ] This yields the solutions operatorname{cis} 12^circ, operatorname{cis} 132^circ, operatorname{cis} 252^circ, ldots continuing every 120^circ for 10 solutions in this case. Combining and sorting these solutions, calculate theta_{2} + theta_{4} + dots + theta_{2n}, and we get the sum as boxed{2280^circ}.