answer:Firstly, notice that: [ (sqrt{2007}+sqrt{2008})(sqrt{2007}-sqrt{2008}) = sqrt{2007}^2 - sqrt{2008}^2 = 2007 - 2008 = -1 ] Additionally: [ (-sqrt{2007}-sqrt{2008})(-sqrt{2008}+sqrt{2007}) = (sqrt{2008}-sqrt{2007})(sqrt{2008}+sqrt{2007}) = sqrt{2008}^2 - sqrt{2007}^2 = 2008 - 2007 = 1 ] Multiplying these results together: [ (sqrt{2007}+sqrt{2008})(sqrt{2007}-sqrt{2008})(-sqrt{2007}-sqrt{2008})(-sqrt{2008}+sqrt{2007}) = (-1)(1) = -1 ] Thus, we find: [ PQRS = boxed{-1} ]

answer:Given that in the right-angled isosceles triangle ( triangle ABC ), we have ( angle ABC = 90^circ ). The coordinates of points ( A ) and ( B ) are given by ( A(1,0) ) and ( B(3,1) ), respectively. We are required to find the coordinates of vertex ( C ). To find the coordinates of point ( C(x, y) ), we use the fact that in a right-angled isosceles triangle, the lengths of the legs are equal. 1. **Distance Calculation**: - Distance ( AB ): [ AB = sqrt{(3 - 1)^2 + (1 - 0)^2} = sqrt{2^2 + 1^2} = sqrt{4 + 1} = sqrt{5} ] - Since ( triangle ABC ) is isosceles right-angled at ( B ): [ AC = BC = AB cdot sqrt{2} = sqrt{5} cdot sqrt{2} = sqrt{10} ] 2. **Using Distance Formula**: [ AC^2 = (x - 1)^2 + (y - 0)^2 ] [ BC^2 = (x - 3)^2 + (y - 1)^2 ] 3. **Equating the Distances**: - Since ( AC = BC ): [ sqrt{10} = (x - 1)^2 + y^2 = (x - 3)^2 + (y - 1)^2 ] 4. **Setting up the Equation**: [ (x-1)^2 + y^2 = (x-3)^2 + (y-1)^2 ] - Expanding both sides: [ (x-1)^2 + y^2 = (x-3)^2 + (y-1)^2 x^2 - 2x + 1 + y^2 = x^2 - 6x + 9 + y^2 - 2y + 1 ] - Simplify and solve for (x): [ -2x + 1 = -6x + 10 - 2y 4x - 2y = 9 2x - y = frac{9}{2} ] 5. **Solutions**: - **Case 1**: ( 2x - y = 1 ): [ y = 2x - 1, x = frac{5}{2},~y = 2(frac{5}{2}) - 1 = 4 (2, 3) ] - **Case 2**: RSA(3, 1 cdot rsqrt 2= (x - 4),~y = -1(4, -1) ] Conclusion: Therefore the coordinates of ( C ) could be either ( (2, 3) ) or ( (4, -1) ): [ boxed{(2, 3) text{ or } (4,-1)} ]

answer:Let's assume one piece of the wire is cut to a length of x cm, and the other piece, therefore, has a length of (l-x) cm. According to the problem, the total area S can be expressed as S=S_1+S_2= frac{x}{6} cdot frac{x}{3} + frac{3(l-x)}{10} cdot frac{l-x}{5} = frac{1}{18}x^2+ frac{3}{50}(l-x)^2, where 0<x<l, Differentiating S with respect to x gives S' = frac{1}{9}x - frac{3}{25}(l-x), Setting S'=0 yields x= frac{27}{52}l, When frac{27}{52}l < x < l, S' > 0, indicating that S is increasing; When 0 < x < frac{27}{52}l, S' < 0, indicating that S is decreasing. Therefore, S reaches its minimum value at x= frac{27}{52}l, and this minimum value is frac{3}{104}l^2. Hence, the answer is: boxed{frac{3}{104}l^2}. To solve this, we can assume one piece of the wire is cut to a length of x cm, forming a rectangle with lengths and widths of frac{2x}{6} and frac{x}{6}, respectively; the other piece, having a length of (l-x) cm, forms a rectangle with lengths and widths of frac{3(l-x)}{10} and frac{2(l-x)}{10}, respectively. By calculating the sum of the areas of these two rectangles and then finding the derivative of this sum with respect to x, we can find the minimum value of the sum of the areas. This problem tests the method of finding the maximum or minimum value of a function, emphasizing the use of derivatives to determine the intervals of monotonicity and thus the extremal values, as well as testing simplification and calculation skills. It is considered a medium-difficulty problem.

answer:1. **Define Variables and Given Conditions:** Let's denote: - ( PD = x ), - ( PE = y ), - ( PF = z ). We are given: ( DE = DF = sqrt{7} ) and ( EF = 2 ). 2. **Set Up Equations Using the Given Conditions:** Considering the triangle inequalities and given distances: [ DE = sqrt{PD^2 + PE^2 - PD cdot PE} = sqrt{7}, ] [ DF = sqrt{PD^2 + PF^2 - PD cdot PF} = sqrt{7}, ] [ EF = sqrt{PE^2 + PF^2 - PE cdot PF} = 2. ] Thus, we establish the following system of equations: [ begin{cases} x^2 + y^2 - xy = 7, x^2 + z^2 - xz = 7, y^2 + z^2 - yz = 4. end{cases} ] 3. **Solve for ( x ):** Subtract the first equation from the second: [ x^2 + y^2 - xy - (x^2 + z^2 - xz) = 7 - 7, ] Simplify to get: [ y^2 - z^2 - x(y - z) = 0 implies x = y + z. ] 4. **Substitute ( x = y + z ) into the System:** Substitute ( x = y + z ) into the equations: [ begin{cases} (y+z)^2 + y^2 - (y+z)y = 7, (y+z)^2 + z^2 - (y+z)z = 7, y^2 + z^2 - yz = 4. end{cases} ] Simplify the first equation: [ y^2 + 2yz + z^2 + y^2 - y^2 - yz = 7, ] [ y^2 + yz + z^2 = 7. ] From the third equation: [ y^2 + z^2 - yz = 4. ] 5. **Solve for ( y^2 + z^2 ) and ( yz ):** Add the two simplified equations: [ (y^2 + yz + z^2) + (y^2 + z^2 - yz) = 7 + 4, ] [ 2(y^2 + z^2) = 11 implies y^2 + z^2 = frac{11}{2}. ] Subtract the second simplified equation from the first: [ (y^2 + yz + z^2) - (y^2 + z^2 - yz) = 7 - 4, ] [ 2yz = 3 implies yz = frac{3}{2}. ] 6. **Find the Values of ( x, y, ) and ( z ):** Rewrite ( x ) as: [ x = y + z. ] Solve for ( y ) and ( z ): Given the quadratic: [ t^2 - (y+z)t + yz = 0 implies t^2 - frac{sqrt{34}}{2}t + frac{3}{2} = 0, ] Solve this quadratic equation for roots ( t ) (where ( t = y, z )): [ t = frac{frac{sqrt{34}}{2} pm sqrt{left(frac{sqrt{34}}{2}right)^2 - 4 cdot frac{3}{2}}}{2}, ] [ y = frac{sqrt{34} + sqrt{10}}{4}, quad z = frac{sqrt{34} - sqrt{10}}{4}. ] Thus, [ x = y + z = frac{sqrt{34}}{2}. ] 7. **Calculate the Volume of the Tetrahedron ( P-DEF ):** The volume of a tetrahedron with vertices including the origin is given by: [ V = frac{1}{6} left| mathbf{a} cdot (mathbf{b} times mathbf{c}) right|. ] For a regular tetrahedron, [ V = frac{sqrt{2}}{12} cdot a^3. ] Therefore, the ratio of the volumes is: [ frac{V_{P-DEF}}{V_{P-ABC}} = frac{xyz}{a^3}, ] Substituting ( xyz ): [ V_{P-DEF} = frac{sqrt{2}}{12} cdot frac{sqrt{34}}{2} cdot frac{3}{2} = frac{sqrt{17}}{8}. ] # Conclusion: [ boxed{frac{sqrt{17}}{8}} ]