💡Okay, so I have this problem about converting parametric equations to polar coordinates and finding their intersection points. Let me try to figure this out step by step.First, the problem gives me a line l with parametric equations:[begin{cases}x = t y = 4 - tend{cases}]where t is a parameter. I need to express these in polar coordinates. I remember that in polar coordinates, x is ρcosθ and y is ρsinθ. So maybe I can substitute these into the parametric equations?Let me try that. If x = t, then t = ρcosθ. Similarly, y = 4 - t, so substituting t from the first equation, y = 4 - ρcosθ. But y is also ρsinθ. So I have:[rho sin theta = 4 - rho cos theta]Let me rearrange this equation to get all the ρ terms on one side:[rho sin theta + rho cos theta = 4]Factor out ρ:[rho (sin theta + cos theta) = 4]So, solving for ρ, I get:[rho = frac{4}{sin theta + cos theta}]Hmm, that seems reasonable. So the polar equation of the line l is ρ = 4 / (sinθ + cosθ). I think that's the answer for part 1.Now, moving on to part 2. I need to find the intersection points of line l and curve C. Curve C has the polar equation ρ = 4 cosθ. So I have two equations:1. ρ = 4 / (sinθ + cosθ)2. ρ = 4 cosθTo find the intersection points, I can set these two equations equal to each other:[frac{4}{sin theta + cos theta} = 4 cos theta]Let me simplify this. Divide both sides by 4:[frac{1}{sin theta + cos theta} = cos theta]Multiply both sides by (sinθ + cosθ):[1 = cos theta (sin theta + cos theta)]Expand the right side:[1 = cos theta sin theta + cos^2 theta]Hmm, I remember that cos^2θ can be written as (1 + cos2θ)/2, and sinθ cosθ is (sin2θ)/2. Maybe I can use these identities to simplify.So, substituting:[1 = frac{sin 2theta}{2} + frac{1 + cos 2theta}{2}]Combine the terms:[1 = frac{sin 2theta + 1 + cos 2theta}{2}]Multiply both sides by 2:[2 = sin 2theta + 1 + cos 2theta]Subtract 1 from both sides:[1 = sin 2theta + cos 2theta]Now, this looks like a single trigonometric function. I remember that a sinx + b cosx can be written as R sin(x + φ), where R = sqrt(a² + b²) and tanφ = b/a. In this case, a = 1 and b = 1, so R = sqrt(2) and φ = 45 degrees or π/4 radians.So, sin2θ + cos2θ = sqrt(2) sin(2θ + π/4). Therefore:[1 = sqrt{2} sinleft(2theta + frac{pi}{4}right)]Divide both sides by sqrt(2):[frac{1}{sqrt{2}} = sinleft(2theta + frac{pi}{4}right)]I know that sin(π/4) = 1/√2, so:[sinleft(2theta + frac{pi}{4}right) = sinleft(frac{pi}{4}right)]This implies that:[2theta + frac{pi}{4} = frac{pi}{4} + 2kpi quad text{or} quad 2theta + frac{pi}{4} = pi - frac{pi}{4} + 2kpi]Simplify the first equation:[2theta = 0 + 2kpi implies theta = kpi]But since 0 ≤ θ < 2π, the solutions are θ = 0 and θ = π.Simplify the second equation:[2theta + frac{pi}{4} = frac{3pi}{4} + 2kpi implies 2theta = frac{pi}{2} + 2kpi implies theta = frac{pi}{4} + kpi]Again, within 0 ≤ θ < 2π, the solutions are θ = π/4 and θ = 5π/4.Now, let's check each θ to see if they satisfy the original equation.First, θ = 0:Plug into ρ = 4 cosθ: ρ = 4 cos0 = 4*1 = 4Check in the other equation: ρ = 4 / (sin0 + cos0) = 4 / (0 + 1) = 4. So it works.Next, θ = π:ρ = 4 cosπ = 4*(-1) = -4. But ρ must be ≥ 0, so this is invalid.Next, θ = π/4:ρ = 4 cos(π/4) = 4*(√2/2) = 2√2Check in the other equation: ρ = 4 / (sin(π/4) + cos(π/4)) = 4 / (√2/2 + √2/2) = 4 / (√2) = 2√2. So it works.Next, θ = 5π/4:ρ = 4 cos(5π/4) = 4*(-√2/2) = -2√2. Again, ρ is negative, so invalid.So the valid solutions are θ = 0 and θ = π/4.Therefore, the polar coordinates of the intersection points are (4, 0) and (2√2, π/4).I think that's it. Let me just recap:1. Converted the parametric equations to polar form by substituting x and y with ρcosθ and ρsinθ, then solved for ρ.2. Set the two polar equations equal to each other to find θ, solved the resulting trigonometric equation, found θ = 0 and θ = π/4.3. Plugged these θ values back into one of the equations to find ρ, ensuring ρ is non-negative.4. Discarded the solutions where ρ was negative.Everything seems to check out. I don't think I made any mistakes, but it's always good to double-check the trigonometric steps, especially when dealing with identities and multiple angles.**Final Answer**1. The polar coordinate equation of the line ( l ) is (boxed{rho(costheta + sintheta) = 4}).2. The polar coordinates of the intersection points are (boxed{(4, 0)}) and (boxed{left(2sqrt{2}, frac{pi}{4}right)}).

💡First, I need to determine how many hours Mark has studied in total over the first four weeks. Adding up the hours from each week: 10 + 14 + 9 + 13 equals 46 hours.Mark's goal is to study an average of 12 hours per week over 5 weeks. To find the total number of hours he needs to study, I multiply the average by the number of weeks: 12 hours/week * 5 weeks = 60 hours.To find out how many hours Mark needs to study in the fifth week, I subtract the total hours he has already studied from his goal: 60 hours - 46 hours = 14 hours.Therefore, Mark must study 14 hours in the fifth week to achieve his goal.

💡Okay, so I have this problem about a geometric sequence. Let me try to understand what it's asking. There's a geometric sequence {a_n} where all the terms are positive. They tell me that a_2 is 4 and a_4 is 16. First, I need to find the common ratio q. Hmm, in a geometric sequence, each term is the previous term multiplied by the common ratio q. So, a_2 is a_1 times q, and a_4 is a_1 times q cubed. Let me write down what I know:- a_2 = a_1 * q = 4- a_4 = a_1 * q^3 = 16I think I can set up these two equations and solve for q. Maybe if I divide the second equation by the first, I can eliminate a_1. Let me try that.So, if I take a_4 divided by a_2, that would be (a_1 * q^3) / (a_1 * q) = 16 / 4. Simplifying that, the a_1 cancels out, and I get q^2 = 4. Wait, so q squared is 4. That means q could be 2 or -2. But the problem says all terms are positive, so q must be positive. Therefore, q is 2. That seems straightforward.Alright, part (1) is done. Now, part (2) says that a_3 and a_5 are the 3rd and 5th terms of an arithmetic sequence {b_n}. I need to find the general term formula for {b_n}.Let me recall that in an arithmetic sequence, each term is the previous term plus a common difference d. So, if I know two terms, I can find d by subtracting them and dividing by the number of intervals between them.First, I need to find a_3 and a_5 from the geometric sequence. Since I know q is 2, let's compute those.a_3 is a_1 * q^2. But I don't know a_1 yet. Wait, from a_2 = 4, which is a_1 * q = 4. Since q is 2, a_1 is 4 / 2 = 2. So, a_1 is 2. Then, a_3 = 2 * 2^2 = 2 * 4 = 8. Similarly, a_5 = 2 * 2^4 = 2 * 16 = 32.So, in the arithmetic sequence {b_n}, the 3rd term is 8 and the 5th term is 32. Let me write that down:- b_3 = 8- b_5 = 32Now, to find the common difference d. The number of intervals between the 3rd and 5th term is 2 (from 3 to 4, and 4 to 5). So, d = (b_5 - b_3) / (5 - 3) = (32 - 8) / 2 = 24 / 2 = 12.So, the common difference d is 12. Now, I need to find the general term formula for {b_n}. In an arithmetic sequence, the nth term is given by b_n = b_1 + (n - 1)d. But I don't know b_1 yet. I know b_3 = 8, which is b_1 + 2d. Since d is 12, let's plug that in.b_3 = b_1 + 2*12 = b_1 + 24 = 8. So, solving for b_1: b_1 = 8 - 24 = -16.Wait, that seems a bit odd. The first term is negative? But the geometric sequence had all positive terms, but the arithmetic sequence isn't necessarily related except for those two terms. So, maybe it's okay.So, the general term is b_n = -16 + (n - 1)*12. Let me simplify that.First, distribute the 12: b_n = -16 + 12n - 12. Combine like terms: -16 -12 is -28. So, b_n = 12n - 28.Let me check if that makes sense. For n=3, b_3 should be 12*3 -28 = 36 -28 = 8. That's correct. For n=5, b_5 = 12*5 -28 = 60 -28 = 32. That's also correct. So, it seems like the formula works. Therefore, the general term for {b_n} is 12n -28.Wait, let me just think again if I did everything correctly. I found q by dividing a_4 by a_2, which gave me q squared equals 4, so q is 2. Then, I found a_1 by dividing a_2 by q, which was 2. Then, a_3 is 8 and a_5 is 32. For the arithmetic sequence, I found the common difference by subtracting 8 from 32 and dividing by 2, which gave me 12. Then, using the formula for the nth term, I solved for b_1 and got -16. Plugging back into the formula, I got 12n -28, which checks out for n=3 and n=5. I think that's correct. The negative first term is a bit unexpected, but since the arithmetic sequence only specifies terms 3 and 5, it's possible for the earlier terms to be negative. So, I think my answer is right.**Final Answer**(1) The common ratio is boxed{2}.(2) The general term formula for the arithmetic sequence is boxed{12n - 28}.

💡First, I need to understand the problem. We have two sets, M and N. M is given as {-1, 0, 1}, and N is defined as {a, a²}. The condition provided is that the intersection of M and N equals N, which implies that N is a subset of M.To satisfy this condition, every element in N must also be in M. Therefore, both a and a² must be elements of M. Since M contains only -1, 0, and 1, I should consider the possible values of a that make both a and a² part of M.Let's analyze the possible values for a:1. If a = -1: - a = -1, which is in M. - a² = (-1)² = 1, which is also in M. - Both elements of N are in M, so N is a subset of M.2. If a = 0: - a = 0, which is in M. - a² = 0² = 0, which is also in M. - Both elements of N are in M, so N is a subset of M.3. If a = 1: - a = 1, which is in M. - a² = 1² = 1, which is also in M. - Both elements of N are in M, so N is a subset of M.4. If a = 2: - a = 2, which is not in M. - a² = 2² = 4, which is also not in M. - Neither element of N is in M, so N is not a subset of M.5. If a = -2: - a = -2, which is not in M. - a² = (-2)² = 4, which is also not in M. - Neither element of N is in M, so N is not a subset of M.From this analysis, the valid values for a that satisfy the condition are -1, 0, and 1. Therefore, the real numbers a that make N a subset of M are -1, 0, and 1.

💡Alright, so I have this problem in the space rectangular coordinate system Oxyz. There are two points given: A(0, a, 1) and B(-1, 2, 2). The problem has two parts. First, I need to find all real numbers a such that triangle OAB is an acute triangle. Second, I need to find the value of a that minimizes the area of triangle OAB. Let me start with part (1). To determine when triangle OAB is acute, I remember that in a triangle, if all angles are less than 90 degrees, then it's an acute triangle. For a triangle to be acute, the square of each side must be less than the sum of the squares of the other two sides. So, I can use the dot product to check the angles because if the dot product of two vectors is positive, the angle between them is acute.First, I need to find the vectors OA, OB, and AB. Vector OA is from the origin O(0,0,0) to A(0,a,1), so OA = (0, a, 1).Vector OB is from O(0,0,0) to B(-1,2,2), so OB = (-1, 2, 2).Vector AB is from A(0,a,1) to B(-1,2,2), so AB = B - A = (-1 - 0, 2 - a, 2 - 1) = (-1, 2 - a, 1).Now, to check the angles at each vertex, I need to compute the dot products of the vectors at each angle.First, let's check the angle at O. The vectors forming this angle are OA and OB. The dot product of OA and OB is:OA · OB = (0)(-1) + (a)(2) + (1)(2) = 0 + 2a + 2 = 2a + 2.For the angle at O to be acute, this dot product must be positive:2a + 2 > 0 ⇒ 2a > -2 ⇒ a > -1.Okay, so a must be greater than -1.Next, let's check the angle at A. The vectors forming this angle are AO and AB. Vector AO is just -OA = (0, -a, -1). Vector AB is (-1, 2 - a, 1). So, the dot product AO · AB is:(0)(-1) + (-a)(2 - a) + (-1)(1) = 0 - 2a + a² - 1 = a² - 2a - 1.For the angle at A to be acute, this dot product must be positive:a² - 2a - 1 > 0.This is a quadratic inequality. Let's solve the equation a² - 2a - 1 = 0.Using the quadratic formula: a = [2 ± sqrt(4 + 4)] / 2 = [2 ± sqrt(8)] / 2 = [2 ± 2√2] / 2 = 1 ± √2.So, the inequality a² - 2a - 1 > 0 holds when a < 1 - √2 or a > 1 + √2.But from the first condition, a > -1. So, combining these, for a > 1 + √2 or a < 1 - √2, but since a > -1, the valid intervals are a > 1 + √2 or -1 < a < 1 - √2.Wait, but 1 - √2 is approximately 1 - 1.414 ≈ -0.414, which is greater than -1. So, the interval -1 < a < 1 - √2 is valid.Now, let's check the angle at B. The vectors forming this angle are BO and BA. Vector BO is -OB = (1, -2, -2). Vector BA is A - B = (0 - (-1), a - 2, 1 - 2) = (1, a - 2, -1). So, the dot product BO · BA is:(1)(1) + (-2)(a - 2) + (-2)(-1) = 1 - 2a + 4 + 2 = 7 - 2a.For the angle at B to be acute, this dot product must be positive:7 - 2a > 0 ⇒ -2a > -7 ⇒ a < 7/2.So, combining all three conditions:1. a > -12. a > 1 + √2 or a < 1 - √23. a < 7/2So, let's see. The intersection of these conditions.From condition 1: a > -1.From condition 2: a > 1 + √2 or a < 1 - √2.But since a > -1, the a < 1 - √2 is only valid if 1 - √2 > -1. Let's check: 1 - √2 ≈ -0.414, which is greater than -1, so the interval -1 < a < 1 - √2 is valid.From condition 3: a < 7/2, which is 3.5.So, combining all:Either a > 1 + √2 and a < 7/2, which is 1 + √2 ≈ 2.414 < a < 3.5.Or a < 1 - √2 and a > -1, which is -1 < a < 1 - √2 ≈ -0.414.But wait, in the angle at A, we had a² - 2a - 1 > 0, which is a > 1 + √2 or a < 1 - √2. So, for the triangle to be acute, all three angles must be acute, so all three conditions must hold.Therefore, the set of a is the intersection of all three conditions:Either:-1 < a < 1 - √2, but we need to ensure that in this interval, all three conditions hold.Wait, but when a is between -1 and 1 - √2, which is approximately -0.414, does the angle at B still hold? Because a < 1 - √2 is approximately -0.414, which is greater than -1, so a is between -1 and -0.414.In this interval, a < 1 - √2, so the angle at A is acute, and a < 7/2, so the angle at B is acute, and a > -1, so the angle at O is acute.Similarly, when a is between 1 + √2 and 7/2, all three conditions hold.So, the set of a is:-1 < a < 1 - √2 or 1 + √2 < a < 7/2.But wait, in the initial step, when I checked the angle at A, I had a² - 2a - 1 > 0, which is a > 1 + √2 or a < 1 - √2.But in the angle at O, we have a > -1, and in the angle at B, a < 7/2.Therefore, the valid intervals are:-1 < a < 1 - √2 and 1 + √2 < a < 7/2.But wait, 1 - √2 is approximately -0.414, so -1 < a < -0.414 and 2.414 < a < 3.5.So, the set of a is:a ∈ (-1, 1 - √2) ∪ (1 + √2, 7/2).But let me verify this because sometimes when dealing with 3D vectors, the triangle inequality might have more conditions.Alternatively, another approach is to compute the lengths of the sides and then apply the condition for an acute triangle, which is that the square of each side is less than the sum of the squares of the other two sides.Let me try this approach as well to confirm.First, compute the lengths of OA, OB, and AB.OA: distance from O to A is sqrt(0² + a² + 1²) = sqrt(a² + 1).OB: distance from O to B is sqrt((-1)² + 2² + 2²) = sqrt(1 + 4 + 4) = sqrt(9) = 3.AB: distance from A to B is sqrt((-1 - 0)² + (2 - a)² + (2 - 1)²) = sqrt(1 + (2 - a)² + 1) = sqrt(2 + (2 - a)²).Now, for triangle OAB to be acute, the following must hold:OA² + OB² > AB²,OA² + AB² > OB²,OB² + AB² > OA².Let me compute each condition.First condition: OA² + OB² > AB².OA² = a² + 1,OB² = 9,AB² = 2 + (2 - a)² = 2 + 4 - 4a + a² = 6 - 4a + a².So, OA² + OB² = (a² + 1) + 9 = a² + 10.Compare with AB² = a² - 4a + 6.So, condition 1: a² + 10 > a² - 4a + 6 ⇒ 10 > -4a + 6 ⇒ 4 > -4a ⇒ -1 < a.Which is the same as before.Second condition: OA² + AB² > OB².OA² = a² + 1,AB² = a² - 4a + 6,OB² = 9.So, OA² + AB² = (a² + 1) + (a² - 4a + 6) = 2a² - 4a + 7.Condition: 2a² - 4a + 7 > 9 ⇒ 2a² - 4a - 2 > 0 ⇒ a² - 2a - 1 > 0.Which is the same as before, leading to a > 1 + √2 or a < 1 - √2.Third condition: OB² + AB² > OA².OB² = 9,AB² = a² - 4a + 6,OA² = a² + 1.So, OB² + AB² = 9 + a² - 4a + 6 = a² - 4a + 15.Compare with OA² = a² + 1.Condition: a² - 4a + 15 > a² + 1 ⇒ -4a + 15 > 1 ⇒ -4a > -14 ⇒ a < 14/4 ⇒ a < 3.5.Which is the same as before.So, combining all three conditions:1. a > -1,2. a > 1 + √2 or a < 1 - √2,3. a < 3.5.Thus, the solution set is:-1 < a < 1 - √2 or 1 + √2 < a < 3.5.So, that's consistent with the previous method.Therefore, the set of all real numbers a that make triangle OAB acute is:a ∈ (-1, 1 - √2) ∪ (1 + √2, 7/2).Now, moving on to part (2): finding the value of a that minimizes the area of triangle OAB.I know that the area of a triangle given by vectors OA and OB is (1/2) |OA × OB|, where × denotes the cross product.Alternatively, since we have three points O, A, B, the area can also be calculated using the formula:Area = (1/2) |AB × AO|, but I think using OA and OB is simpler.So, let's compute the cross product OA × OB.First, OA = (0, a, 1),OB = (-1, 2, 2).The cross product OA × OB is:|i   j   k||0   a   1||-1  2   2|= i*(a*2 - 1*2) - j*(0*2 - 1*(-1)) + k*(0*2 - a*(-1))= i*(2a - 2) - j*(0 + 1) + k*(0 + a)= (2a - 2, -1, a).Now, the magnitude of this cross product is:|OA × OB| = sqrt[(2a - 2)^2 + (-1)^2 + a^2]= sqrt[(4a² - 8a + 4) + 1 + a²]= sqrt[5a² - 8a + 5].Therefore, the area of triangle OAB is (1/2) * sqrt(5a² - 8a + 5).To minimize the area, we need to minimize the expression under the square root, which is 5a² - 8a + 5.This is a quadratic function in terms of a, and since the coefficient of a² is positive, it opens upwards, so the minimum occurs at the vertex.The vertex of a quadratic ax² + bx + c is at a = -b/(2a).Here, a = 5, b = -8.So, the value of a that minimizes the quadratic is:a = -(-8)/(2*5) = 8/10 = 4/5.Therefore, the area is minimized when a = 4/5.Let me verify this by computing the derivative.Let f(a) = 5a² - 8a + 5.f'(a) = 10a - 8.Setting f'(a) = 0:10a - 8 = 0 ⇒ a = 8/10 = 4/5.Yes, that confirms it.So, the value of a that minimizes the area is 4/5.Therefore, the answers are:(1) The set of a is (-1, 1 - √2) ∪ (1 + √2, 7/2).(2) The value of a that minimizes the area is 4/5.

💡Okay, so I have this expression to simplify:[frac{left(x^{2 / m}-9 x^{2 / n}right)left(sqrt[m]{x^{1-m}}-3 sqrt[n]{x^{1-n}}right)}{left(x^{1 / m}+3 x^{1 / n}right)^{2}-12 x^{(m+n) /(m n)}}]Hmm, that looks pretty complicated, but maybe I can break it down step by step.First, let me look at the numerator:[left(x^{2 / m}-9 x^{2 / n}right)left(sqrt[m]{x^{1-m}}-3 sqrt[n]{x^{1-n}}right)]I notice that (x^{2/m} - 9x^{2/n}) looks like a difference of squares. Maybe I can factor that. Let me try:[x^{2/m} - 9x^{2/n} = left(x^{1/m}right)^2 - left(3x^{1/n}right)^2 = left(x^{1/m} - 3x^{1/n}right)left(x^{1/m} + 3x^{1/n}right)]Okay, so that part factors nicely. Now, the other part of the numerator is:[sqrt[m]{x^{1 - m}} - 3sqrt[n]{x^{1 - n}}]I can rewrite the roots as exponents:[sqrt[m]{x^{1 - m}} = x^{(1 - m)/m} = x^{1/m - 1}][sqrt[n]{x^{1 - n}} = x^{(1 - n)/n} = x^{1/n - 1}]So, the second part becomes:[x^{1/m - 1} - 3x^{1/n - 1}]Hmm, maybe I can factor out (x^{-1}) from both terms:[x^{-1}left(x^{1/m} - 3x^{1/n}right)]Yes, that works. So, the entire numerator now is:[left(x^{1/m} - 3x^{1/n}right)left(x^{1/m} + 3x^{1/n}right) cdot x^{-1}left(x^{1/m} - 3x^{1/n}right)]Wait, that's:[left(x^{1/m} - 3x^{1/n}right)^2 cdot left(x^{1/m} + 3x^{1/n}right) cdot x^{-1}]Okay, so the numerator simplifies to:[x^{-1} left(x^{1/m} - 3x^{1/n}right)^2 left(x^{1/m} + 3x^{1/n}right)]Now, let's look at the denominator:[left(x^{1 / m} + 3 x^{1 / n}right)^{2} - 12 x^{(m + n) / (m n)}]First, expand (left(x^{1/m} + 3x^{1/n}right)^2):[left(x^{1/m}right)^2 + 2 cdot x^{1/m} cdot 3x^{1/n} + left(3x^{1/n}right)^2 = x^{2/m} + 6x^{1/m + 1/n} + 9x^{2/n}]So, the denominator becomes:[x^{2/m} + 6x^{1/m + 1/n} + 9x^{2/n} - 12x^{(m + n)/(mn)}]Hmm, I notice that (1/m + 1/n = (m + n)/(mn)), so (x^{1/m + 1/n} = x^{(m + n)/(mn)}). Therefore, the denominator simplifies to:[x^{2/m} + 6x^{(m + n)/(mn)} + 9x^{2/n} - 12x^{(m + n)/(mn)} = x^{2/m} - 6x^{(m + n)/(mn)} + 9x^{2/n}]Wait a second, that looks like another difference of squares. Let me check:[x^{2/m} - 6x^{(m + n)/(mn)} + 9x^{2/n} = left(x^{1/m} - 3x^{1/n}right)^2]Yes, because:[left(x^{1/m} - 3x^{1/n}right)^2 = x^{2/m} - 6x^{1/m + 1/n} + 9x^{2/n}]Which is exactly what we have. So, the denominator simplifies to:[left(x^{1/m} - 3x^{1/n}right)^2]Now, putting it all together, the original expression becomes:[frac{x^{-1} left(x^{1/m} - 3x^{1/n}right)^2 left(x^{1/m} + 3x^{1/n}right)}{left(x^{1/m} - 3x^{1/n}right)^2}]I can cancel out the (left(x^{1/m} - 3x^{1/n}right)^2) terms in the numerator and denominator:[x^{-1} left(x^{1/m} + 3x^{1/n}right)]Which is the same as:[frac{x^{1/m} + 3x^{1/n}}{x}]So, the simplified expression is:[frac{x^{1/m} + 3x^{1/n}}{x}]I should double-check to make sure I didn't make any mistakes, especially with the exponents. Let me verify each step:1. Factoring the numerator as a difference of squares: Correct.2. Rewriting the roots as exponents: Correct.3. Factoring out (x^{-1}): Correct.4. Simplifying the denominator by expanding and recognizing another difference of squares: Correct.5. Canceling the common terms: Correct.6. Final simplification: Correct.Everything seems to check out. I think this is the correct simplified form.