answer:Given the equation: [ 2x^2 - 6x + y^2 = 0 ] our goal is to find the maximum value of the expression: [ x^2 + y^2 + 2x ] 1. From the given equation, isolate (y^2): [ y^2 = -2x^2 + 6x ] 2. Substitute (y^2) into the expression (x^2 + y^2 + 2x): [ x^2 + y^2 + 2x = x^2 + (-2x^2 + 6x) + 2x ] [ = x^2 - 2x^2 + 6x + 2x ] [ = -x^2 + 8x ] 3. We need to maximize (-x^2 + 8x). It forms a downward-opening parabola, so its maximum value occurs at the vertex. The vertex of (ax^2 + bx + c) is given by (x = -frac{b}{2a}). In our case, (a = -1) and (b = 8): [ x = -frac{8}{2(-1)} = 4 ] 4. Substitute back (x = 4) into the problem to check if it satisfies the original condition (2x^2 - 6x + y^2 = 0): [ 2(4)^2 - 6(4) + y^2 = 0 ] [ 32 - 24 + y^2 = 0 ] [ y^2 = -8 ] Since (y^2 = -8) is not possible (as square of a real number cannot be negative), we need to find the value of (x) within the range that does achieve a valid maximum. 5. The condition for (y^2 geq 0) must hold: [ -2x^2 + 6x geq 0 ] 6. Solve the inequality: [ 2x^2 - 6x leq 0 ] [ x(2x - 6) leq 0 ] [ x(2x - 6) leq 0 ] [ x(2x - 6) leq 0 ] Hence, the roots of the quadratic are (x = 0) and (x = 3). The inequality holds for: [ 0 leq x leq 3 ] 7. To find the maximum value of (-x^2 + 8x) within the interval (0 leq x leq 3): - Evaluate at (x = 0): [ -0^2 + 8(0) = 0 ] - Evaluate at (x = 3): [ -(3)^2 + 8(3) = -9 + 24 = 15 ] So, the maximum value is (15). Thus, Conclusion: [ boxed{15} ]

answer:1. **Understanding the Problem:** We need to find the greatest possible product of the digits of a positive integer ( N ) whose digits sum up to 23. 2. **Strategy:** To maximize the product of the digits, we should use the digits in such a way that their product is maximized. Generally, the digit 9 is the most efficient in terms of maximizing the product, but we need to ensure the sum of the digits is 23. 3. **Using Digits Efficiently:** - The digit 9 is the largest single-digit number and will contribute significantly to the product. - We need to find a combination of digits that sum to 23 and maximize the product. 4. **Initial Calculation:** - If we use two 9's, their sum is ( 9 + 9 = 18 ). - We need additional digits that sum to ( 23 - 18 = 5 ). 5. **Finding the Remaining Digits:** - The remaining sum is 5. The digits that sum to 5 and maximize the product are 5 itself (since 5 is greater than any combination of smaller digits that sum to 5). 6. **Combining the Digits:** - Using two 9's and one 5, the digits are ( 9, 9, ) and ( 5 ). - The product of these digits is ( 9 times 9 times 5 = 81 times 5 = 405 ). 7. **Verification:** - The sum of the digits ( 9 + 9 + 5 = 23 ). - The product is ( 405 ). 8. **Exploring Other Combinations:** - If we use one 9, the remaining sum is ( 23 - 9 = 14 ). - The digits that sum to 14 and maximize the product are ( 8 + 6 ) (since ( 8 times 6 = 48 ) is greater than any other combination of digits summing to 14). 9. **Combining the Digits:** - Using one 9, one 8, and one 6, the digits are ( 9, 8, ) and ( 6 ). - The product of these digits is ( 9 times 8 times 6 = 72 times 6 = 432 ). 10. **Verification:** - The sum of the digits ( 9 + 8 + 6 = 23 ). - The product is ( 432 ). 11. **Conclusion:** - The combination ( 9, 8, 6 ) provides a higher product than ( 9, 9, 5 ). The final answer is (boxed{432})

answer:# Problem: 例 1 若关于 x 的方程 2^{2x} + 2^{x} a + a + 1 = 0 有实数根，则实数 a 的取值范围是： To determine the values of a for which the equation 2^{2x} + 2^{x} a + a + 1 = 0 has real roots, we start by analyzing the equation. 1. Introduce a substitution. Let ( y = 2^x ). Then ( 2^{2x} = y^2 ). Substituting in the original equation, we get: [ y^2 + y a + a + 1 = 0 ] 2. This is a quadratic equation in ( y ). To have real roots ( y ), the discriminant must be non-negative. The discriminant (Delta ) of the quadratic equation ( y^2 + y a + a + 1 = 0 ) is given by: [ Delta = a^2 - 4(a + 1) ] 3. For real roots to exist, the discriminant must be non-negative: [ a^2 - 4(a + 1) geq 0 ] 4. Simplify the inequality: [ a^2 - 4a - 4 geq 0 ] 5. Solve this quadratic inequality. First, find the roots of the quadratic equation ( a^2 - 4a - 4 = 0 ) using the quadratic formula ( a = frac{-b pm sqrt{b^2 - 4ac}}{2a} ): [ a = frac{4 pm sqrt{16 + 16}}{2} = frac{4 pm 4sqrt{2}}{2} ] [ a = 2 pm 2sqrt{2} ] 6. The roots are ( 2 + 2sqrt{2} ) and ( 2 - 2sqrt{2} ), so the quadratic equation can be factored as: [ (a - (2 + 2sqrt{2}))(a - (2 - 2sqrt{2})) geq 0 ] 7. To solve the inequality ( (a - 2 - 2sqrt{2})(a - 2 + 2sqrt{2}) geq 0 ), consider the intervals determined by the roots: [ (-infty, 2 - 2sqrt{2}] cup [2 + 2sqrt{2}, +infty) ] These intervals are where the product of the factors is non-negative. # Conclusion: For the given quadratic equation ( 2^{2x} + 2^x a + a + 1 = 0 ) in terms of ( y = 2^x ) to have real roots, the value of ( a ) must lie in: [ 2 + 2sqrt{2} leq a leq 2 - 2sqrt{2} ] Therefore, the range of ( a ) is: [ boxed{(-infty, 2 - 2sqrt{2}] cup [2 + 2sqrt{2}, +infty)} ]

answer:1. **Representation in the given form**: We need to show that every positive integer ( N ) can be uniquely represented as [ N = sum_{i=0}^k d_i 2^i, ] where ( k ) is a nonnegative integer and each ( d_i ) is either 1 or 2. 2. **Transformation to binary-like form**: Consider the representation of ( N ) in a form similar to binary notation but with digits 1 and 2. Let us rewrite ( N ) as: [ N = sum_{i=0}^k (e_i + 1) 2^i, ] where ( e_i ) is either 0 or 1. This transformation is valid because ( d_i ) can be either 1 or 2, and thus ( d_i = e_i + 1 ). 3. **Simplifying the expression**: We can simplify the above expression: [ N = sum_{i=0}^k e_i 2^i + sum_{i=0}^k 2^i. ] The second sum is a geometric series: [ sum_{i=0}^k 2^i = 2^{k+1} - 1. ] Therefore, we have: [ N = sum_{i=0}^k e_i 2^i + (2^{k+1} - 1). ] 4. **Unique representation in binary**: We need to prove the unique representation of the number ( N - 2^{k+1} + 1 ) in the form ( sum_{i=0}^k e_i 2^i ). This is equivalent to proving the unique binary representation of ( N - 2^{k+1} + 1 ), which is well-known to be true by the fundamental theorem of arithmetic for binary numbers. 5. **Determining the value of ( k )**: We need to show that there exists only one ( k ) for which this representation is possible. We know that: [ k < log_2(N+1) - 1. ] For ( k = lfloor log_2(N+1) rfloor - 1 ), we receive the correct representation. Now, let's try ( k leq lfloor log_2(N+1) rfloor - 2 ). Then: [ 2^{k+2} < N + 1, ] so: [ N - 2^{k+1} + 1 geq 2^{k+1}. ] Therefore, there isn't any representation of this number in the form ( sum_{i=0}^k e_i 2^i ) because: [ maxleft(sum_{i=0}^k e_i 2^iright) = 2^{k+1} - 1. ] 6. **Conclusion**: Since ( k = lfloor log_2(N+1) rfloor - 1 ) is the only value that satisfies the conditions, the representation is unique. (blacksquare)