answer:Let ( n ) be a positive integer. We start by expressing the numbers ( A ) and ( B ) given the conditions: 1. ( A ) is a ( 2n )-digit number where each digit is 4. Thus, we can write ( A ) in the form: [ A = frac{4}{9} times underbrace{99 cdots 9}_{2n text{ digits}} = frac{4}{9} times (10^{2n} - 1) ] 2. ( B ) is an ( n )-digit number where each digit is 8. Hence, ( B ) can be written as: [ B = frac{8}{9} times underbrace{99 cdots 9}_{n text{ digits}} = frac{8}{9} times (10^n - 1) ] Next, we want to show that ( A + 2B + 4 ) is a perfect square. 3. Substitute the expressions for ( A ) and ( B ) to find ( A + 2B + 4 ): [ A + 2B + 4 = frac{4}{9} times (10^{2n} - 1) + frac{16}{9} times (10^n - 1) + 4 ] 4. Simplify the expression step-by-step: begin{align*} A + 2B + 4 &= frac{4}{9} times 10^{2n} - frac{4}{9} + frac{16}{9} times 10^n - frac{16}{9} + 4 &= frac{4}{9} times 10^{2n} + frac{16}{9} times 10^n - frac{4}{9} - frac{16}{9} + 4 end{align*} Now combine the constant terms: [ = frac{4}{9} times 10^{2n} + frac{16}{9} times 10^n + left(4 - frac{4}{9} - frac{16}{9}right) ] Simplify the constant term: [ 4 - frac{4}{9} - frac{16}{9} = 4 - frac{20}{9} = 4 - frac{20}{9} = frac{36}{9} - frac{20}{9} = frac{16}{9} ] So we get: [ A + 2B + 4 = frac{4}{9} times 10^{2n} + frac{16}{9} times 10^n + frac{16}{9} ] 5. Notice we can factor this expression as: [ A + 2B + 4 = frac{4}{9} left(10^{2n} + 4 times 10^n + 4 right) ] 6. Observe that the terms inside the parentheses form a perfect square: [ 10^{2n} + 4 times 10^n + 4 = (10^n + 2)^2 ] So, [ frac{4}{9} left( (10^n + 2)^2 right) = left( frac{2}{3} (10^n + 2) right)^2 ] 7. Finally, check that ( frac{2}{3} (10^n + 2) ) is an integer: [ 10^n equiv 1 pmod{3} quad text{(since 10 ≡ 1 mod 3)} ] [ 10^n + 2 equiv 1 + 2 equiv 0 pmod{3} ] Thus, ( 10^n + 2 ) is divisible by 3, making ( frac{2}{3} (10^n + 2) ) an integer. Therefore, ( A + 2B + 4 ) is a perfect square. [ boxed{text{Hence, } A + 2B + 4 text{ is a perfect square.}} ]

answer:(1) The domain of f(x) is {x|x > 0}, f′(x)= frac{ax-1}{x}. Let f′(x) > 0, we get x > frac{1}{a}. Let f′(x) < 0, we get 0 < x < frac{1}{a}. Thus, f(x) is decreasing on (0, frac{1}{a}) and increasing on (frac{1}{a},+infty). Thus, the minimum value of f(x) is f(frac{1}{a})=1+ln a=2. Solving for a, we get a=e, hence x_0=frac{1}{a}=frac{1}{e}. (2) Since f(x_1)+f(x_2)-2f(frac{x_1+x_2}{2}) =(ax_1-ln x_1)+(ax_2-ln x_2)-2(afrac{x_1+x_2}{2}-lnfrac{x_1+x_2}{2}) =ln(frac{x_1+x_2}{2})^2-ln x_1x_2 =lnfrac{x_1^2+x_2^2+2x_1x_2}{4x_1x_2} geqslantlnfrac{2x_1x_2+2x_1x_2}{4x_1x_2}=0, Thus, f(x_1)+f(x_2)geqslant 2f(frac{x_1+x_2}{2}). In markdown format, the final answers are: (1) boxed{a=e, x_0=frac{1}{e}} (2) boxed{f(x_1)+f(x_2)geqslant 2f(frac{x_1+x_2}{2})}

answer:Given that (a, b, c) are the roots of the cubic equation: [ x^3 - x^2 - x - 1 = 0 ] we want to show: 1. (a, b, c) are distinct. 2. The value of the algebraic expression (frac{a^{1993}-b^{1993}}{a-b} + frac{b^{1993}-c^{1993}}{b-c} + frac{c^{1993}-a^{1993}}{c-a}) is an integer. # Step-by-Step Solution (1) (a, b, c) are distinct Let's omit part (1), as per the instruction. (2) Proving the expression is an integer Define the function: [ f(n) = frac{a^n - b^n}{a - b} + frac{b^n - c^n}{b - c} + frac{c^n - a^n}{c - a} ] We must show that (f(n)) is an integer for all non-negative integers (n). # Base Cases: 1. **When (n = 0):** [ f(0) = frac{a^0 - b^0}{a - b} + frac{b^0 - c^0}{b - c} + frac{c^0 - a^0}{c - a} = frac{1 - 1}{a - b} + frac{1 - 1}{b - c} + frac{1 - 1}{c - a} = 0 ] 2. **When (n = 1):** [ f(1) = frac{a - b}{a - b} + frac{b - c}{b - c} + frac{c - a}{c - a} = 1 + 1 + 1 = 3 ] These values, (f(0) = 0) and (f(1) = 3), are integers. 3. **When (n = 2):** [ f(2) = frac{a^2 - b^2}{a - b} + frac{b^2 - c^2}{b - c} + frac{c^2 - a^2}{c - a} ] Using the property: [ frac{a^2 - b^2}{a - b} = a + b quad text{(and similarly for others)}, ] we get: [ f(2) = (a + b) + (b + c) + (c + a) = 2(a + b + c) ] Given that (a + b + c = 1) (by Vieta’s formulas), we have: [ f(2) = 2 cdot 1 = 2 ] (f(2) = 2) is also an integer. # Inductive Step: Assume that (f(k - 2)), (f(k - 1)), and (f(k)) are integers for some integer (k geq 2). We need to show that (f(k + 1)) is also an integer. Given that (a, b, c) satisfy the cubic equation: [ a^3 = a^2 + a + 1, quad b^3 = b^2 + b + 1, quad c^3 = c^2 + c + 1, ] we can write: [ a^{k+1} = a cdot a^k = a(a^{k-1} + a^{k-2} + a^{k-3}), ] and similarly for (b) and (c), we obtain: [ f(k+1) = f(k) + f(k-1) + f(k-2) ] By the principle of mathematical induction, if (f(k)), (f(k-1)), and (f(k-2)) are integers, then (f(k+1)) is also an integer. Therefore, since (f(0)), (f(1)), and (f(2)) are integers and the recurrence relation (f(k+1) = f(k) + f(k-1) + f(k-2)) maintains this property, we conclude that: [ f(n) text{ is an integer for all non-negative integers } n. ] Specifically, (f(1993)) is an integer. # Conclusion [ boxed{text{f(n)}} ]

answer:To find the area of an isosceles trapezoid, we can use the formula: Area = (1/2) * (sum of the lengths of the two bases) * height First, we need to find the height of the trapezoid. Since the trapezoid is isosceles, the legs (non-parallel sides) are equal in length, and the height will form two right triangles within the trapezoid, one on each side. Let's call the height "h". We can use the Pythagorean theorem to find the height, since we have a right triangle formed by the height, half the difference of the bases, and the leg of the trapezoid. The difference between the two bases is 13 - 7 = 6. Since the trapezoid is symmetrical, this difference is split into two parts, one for each of the right triangles. So each part is 6/2 = 3 units long. Now we have a right triangle with one leg of length 3 (half the difference of the bases) and the hypotenuse of length 5 (the leg of the trapezoid). We can use the Pythagorean theorem to solve for the height "h": h^2 + 3^2 = 5^2 h^2 + 9 = 25 h^2 = 25 - 9 h^2 = 16 h = √16 h = 4 Now that we have the height, we can find the area of the trapezoid: Area = (1/2) * (7 + 13) * 4 Area = (1/2) * 20 * 4 Area = 10 * 4 Area = 40 The area of the isosceles trapezoid is boxed{40} square units.