🧠:<think>Okay, so I've got this inequality: 2x + 2/(x - a) ≥ 7, and it's supposed to hold true for all x in the interval (a, +∞). I need to find the minimum value of the real number a. Hmm, let me try to figure this out step by step.First, I know that since x is greater than a, the expression (x - a) is positive. That means I don't have to worry about the denominator being zero or negative, which is good because it keeps things simpler.Now, the inequality is 2x + 2/(x - a) ≥ 7. I want this to be true for all x > a, so maybe I can find the minimum value of the left-hand side expression and set that to be at least 7. That way, if the minimum is 7, then all other values will naturally be greater than or equal to 7.Let me denote y = 2x + 2/(x - a). I need to find the minimum value of y. To do this, I can use calculus or maybe the AM-GM inequality. Let me try the AM-GM approach first because it might be quicker.Looking at the expression y = 2x + 2/(x - a), I notice that it's a sum of two terms: one linear in x and one reciprocal in (x - a). Maybe I can rewrite it to make it fit the AM-GM form.Let me set t = x - a. Since x > a, t will be positive. Then, x = a + t, so substituting back into y:y = 2(a + t) + 2/t = 2a + 2t + 2/t.So now, y = 2a + 2t + 2/t. I can factor out the 2:y = 2a + 2(t + 1/t).Now, the expression t + 1/t is a classic one. I remember that for t > 0, t + 1/t is always greater than or equal to 2, with equality when t = 1. That's from the AM-GM inequality: (t + 1/t)/2 ≥ sqrt(t * 1/t) = 1, so t + 1/t ≥ 2.So, substituting back, y = 2a + 2(t + 1/t) ≥ 2a + 2*2 = 2a + 4.Therefore, the minimum value of y is 2a + 4. Since we want y ≥ 7 for all x > a, this minimum must be at least 7:2a + 4 ≥ 7.Subtracting 4 from both sides:2a ≥ 3.Dividing both sides by 2:a ≥ 3/2.So, the minimum value of a is 3/2. Let me just verify this.If a = 3/2, then the expression becomes y = 2x + 2/(x - 3/2). Let's find its minimum. As before, set t = x - 3/2, so x = 3/2 + t, and y becomes:y = 2*(3/2 + t) + 2/t = 3 + 2t + 2/t.Then, y = 3 + 2(t + 1/t). The minimum of t + 1/t is 2, so the minimum y is 3 + 2*2 = 7, which is exactly what we need. So, when a = 3/2, the inequality holds with equality at t = 1, which is x = 3/2 + 1 = 5/2.Therefore, the minimum value of a is indeed 3/2. I think that makes sense. I don't see any mistakes in my reasoning, so I'm pretty confident about this answer.Final AnswerThe minimum value of the real number ( a ) is (boxed{frac{3}{2}}).</think>

🧠:<think>First, calculate the amount of salt in the original solution by multiplying the total volume by the salt concentration.Next, determine the volume of water that evaporates by taking one fourth of the total volume.Subtract the evaporated water from the original volume to find the remaining solution volume.The amount of salt remains unchanged after evaporation.Add the specified amount of salt to the remaining salt to get the new total salt content.Let ( W ) represent the amount of water added. The new total volume of the solution is the sum of the remaining solution and the added water.Set up an equation based on the new salt concentration and solve for ( W ).Finally, present the calculated amount of water added.</think>

🧠:<think>Okay, so I have this problem to prove that if a and b are positive integers and their product ab is even, then at least one of them must be even. The method I need to use is contradiction. Hmm, contradiction proofs can be tricky, but I think I get the basic idea. You assume the opposite of what you want to prove and then show that this leads to a contradiction.Alright, let's start by understanding the statement. If ab is even, then at least one of a or b is even. To use contradiction, I need to assume the opposite of the conclusion. So, the opposite of "at least one of a or b is even" would be "both a and b are odd." That makes sense because if neither is even, then both must be odd.So, I'll assume that both a and b are odd. Now, how do I show that this leads to a contradiction? Well, if a and b are both odd, then their product ab should be... hmm, what happens when you multiply two odd numbers? I remember that an odd number can be written in the form 2k + 1, where k is an integer. So, let's write a and b like that.Let’s say a = 2m + 1 and b = 2n + 1, where m and n are integers. Now, let's multiply them together:ab = (2m + 1)(2n + 1)Let me expand this out:ab = 2m * 2n + 2m * 1 + 1 * 2n + 1 * 1ab = 4mn + 2m + 2n + 1Hmm, okay, so ab = 4mn + 2m + 2n + 1. I can factor out a 2 from the first three terms:ab = 2(2mn + m + n) + 1So, ab is equal to 2 times some integer plus 1. That means ab is odd because it's one more than an even number. But wait, the original statement says that ab is even. So, if ab is even, but according to our assumption, ab is odd, that's a contradiction!Therefore, our initial assumption that both a and b are odd must be false. So, at least one of a or b has to be even. That makes sense. I think I followed the steps correctly. I assumed the opposite, showed that it leads to a contradiction, and therefore proved the original statement.Let me double-check to make sure I didn't make any mistakes. I assumed both a and b are odd, expressed them as 2m + 1 and 2n + 1, multiplied them out, and showed that ab is odd. But the problem states that ab is even, so this is a contradiction. Therefore, my assumption that both are odd is wrong, meaning at least one must be even. Yeah, that seems solid.I guess another way to think about it is that even numbers are multiples of 2. If neither a nor b is even, then neither is a multiple of 2, so their product can't be a multiple of 2 either. But since ab is even, it must be a multiple of 2, which means at least one of a or b has to be a multiple of 2, i.e., even. That reinforces the conclusion.Alright, I think I've got it. Using contradiction, I assumed both a and b are odd, showed that leads to ab being odd, which contradicts the given that ab is even. Therefore, at least one of a or b must be even. Cool, that makes sense!</think>

🧠:<think>Alright, so I have this problem about two skew lines, a and b, which form a 50-degree angle. There's a fixed point P in space, and I need to find how many lines pass through P and form a 30-degree angle with both a and b. The answer is supposed to be exactly 2, but I need to figure out why.First, let me recall what skew lines are. Skew lines are lines in three-dimensional space that do not intersect and are not parallel. So, they lie on different planes. The angle between two skew lines is defined as the angle between their direction vectors when projected onto a common plane. In this case, the angle is 50 degrees.Now, point P is fixed somewhere in space. I need to find lines passing through P that form a 30-degree angle with both a and b. Hmm, okay. So, each such line must make a 30-degree angle with both a and b.I think I can approach this problem by considering the concept of the angle between two lines. The angle between two lines is determined by the angle between their direction vectors. So, if I can find direction vectors for lines passing through P that make 30 degrees with both a and b, that would solve the problem.But how do I visualize this? Maybe I can think of the set of all lines passing through P that make a 30-degree angle with line a. Similarly, the set of all lines passing through P that make a 30-degree angle with line b. The intersection of these two sets would give me the lines that satisfy both conditions.Wait, what does the set of lines through P making a fixed angle with a given line look like? I remember that in three dimensions, the set of lines through a point making a fixed angle with a given line forms a cone. The apex of the cone is at P, and the axis of the cone is the given line. The half-angle at the apex is equal to the fixed angle.So, if I consider line a, the set of lines through P making a 30-degree angle with a forms a cone around a with apex at P and half-angle 30 degrees. Similarly, for line b, it's another cone with the same apex P and half-angle 30 degrees.Therefore, the problem reduces to finding the number of common lines (generators) to both cones. The number of common lines would correspond to the number of solutions, i.e., the number of lines passing through P that make 30 degrees with both a and b.Now, how do two cones intersect? In three-dimensional space, two cones can intersect in 0, 1, or 2 lines, depending on their relative positions and angles. Since the cones here have the same apex, their intersection is more constrained.Given that the original angle between lines a and b is 50 degrees, and we're looking for lines that make 30 degrees with both, I need to see if such lines can exist.Let me think about the angle between the two cones. The angle between the axes of the cones (lines a and b) is 50 degrees. Each cone has a half-angle of 30 degrees. So, the question is: can a line lie on both cones?I remember that for two cones with a common apex, the number of common generators depends on the angle between their axes and their half-angles. There's a formula or a method to determine this.I think it's related to the spherical trigonometry. If I imagine the cones intersecting a unit sphere centered at P, each cone would intersect the sphere in a circle. The angle between the axes is 50 degrees, and each circle has a radius (on the sphere) corresponding to 30 degrees.So, on the unit sphere, we have two circles with angular radius 30 degrees, separated by an angular distance of 50 degrees. The number of intersection points between these two circles will determine the number of common generators.In spherical geometry, two circles can intersect in 0, 1, or 2 points. The number of intersection points depends on the sum and difference of their angular radii and the angular distance between their centers.The formula for the number of intersection points is as follows:If the sum of the radii is less than the angular distance, there are no intersections.If the sum is equal to the angular distance, there's exactly one intersection.If the sum is greater than the angular distance but the difference is less than the angular distance, there are two intersections.If the difference is equal to the angular distance, there's one intersection.If the difference is greater than the angular distance, there are no intersections.Wait, let me recall the exact conditions. I think it's similar to the circle intersection in plane geometry but adapted for spherical geometry.In plane geometry, two circles can intersect in 0, 1, or 2 points depending on the distance between their centers and their radii. Similarly, on a sphere, two circles (which are essentially spherical caps) can intersect in 0, 1, or 2 points.The angular distance between the centers is 50 degrees, and each circle has a radius of 30 degrees.So, let's compute:Sum of radii: 30 + 30 = 60 degrees.Difference of radii: |30 - 30| = 0 degrees.Angular distance between centers: 50 degrees.Now, to determine the number of intersections:If the angular distance is less than the sum and greater than the difference, there are two intersection points.If the angular distance equals the sum or the difference, there's one intersection point.If the angular distance is greater than the sum or less than the difference, there are no intersection points.In our case:Sum of radii: 60 degrees.Angular distance: 50 degrees.Since 50 < 60, and 50 > 0 (which is the difference), there are two intersection points.Therefore, the two circles on the sphere intersect at two points, meaning there are two common generators (lines) passing through P that make 30 degrees with both a and b.Hence, the number of such lines is exactly 2.Wait, but I should verify this reasoning because sometimes the spherical geometry can be a bit tricky.Let me consider the spherical triangle formed by the centers of the two circles and one of their intersection points.The sides of this triangle would be 30 degrees, 30 degrees, and 50 degrees.Using the spherical law of cosines:cos(c) = cos(a)cos(b) + sin(a)sin(b)cos(C)Where a and b are the sides adjacent to angle C, and c is the side opposite angle C.In our case, if we consider the angle between the two radii (30 degrees each) at the intersection point, the angle opposite to the 50-degree side would be something.But actually, I think the key point is that since the sum of the radii (60 degrees) is greater than the angular distance (50 degrees), and the difference is 0, which is less than 50 degrees, so the circles intersect at two points.Therefore, my initial conclusion seems correct.Alternatively, I can think about the problem in terms of vector algebra.Let me denote the direction vectors of lines a and b as u and v, respectively. The angle between u and v is 50 degrees.We need to find a direction vector w such that the angle between w and u is 30 degrees, and the angle between w and v is also 30 degrees.So, mathematically, we have:w ⋅ u = |w||u|cos(30°)w ⋅ v = |w||v|cos(30°)Assuming u and v are unit vectors, and w is also a unit vector for simplicity.So, w ⋅ u = cos(30°)w ⋅ v = cos(30°)We need to find the number of unit vectors w satisfying these two equations.This is equivalent to finding the intersection of two cones in three-dimensional space, which is similar to the earlier geometric approach.The number of solutions depends on the angle between u and v. In our case, the angle is 50 degrees.I recall that if the angle between u and v is greater than twice the angle of the cones (which is 60 degrees in this case), then there are no solutions. If it's equal, there's one solution, and if it's less, there are two solutions.Wait, let me think again.If the angle between u and v is less than the sum of the angles of the cones, then there are two solutions.In our case, the angle between u and v is 50 degrees, which is less than 60 degrees (sum of 30 + 30). Therefore, there are two solutions.Yes, that makes sense.So, algebraically, there are two such direction vectors w, meaning two lines passing through P that make 30 degrees with both a and b.Therefore, the number of such lines is exactly 2.Another way to think about it is by considering the possible orientations of the line through P. Since the angle between a and b is 50 degrees, which is less than 60 degrees (twice 30 degrees), the cones around a and b will intersect in two distinct lines.If the angle between a and b were exactly 60 degrees, the cones would touch at exactly one line. If it were more than 60 degrees, there would be no such lines.But since 50 degrees is less than 60 degrees, we have two lines.I think this confirms the answer.Final AnswerThe number of such lines is exactly boxed{2}.</think>