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中2 幾條數學問題
發問:
1.Factorize 2x^2(y+1)^2-4xy^2(y+1)2.Factorize p^2 r+pr-q^2 r-qr3a.(i)Factorize 2z^3-12z^2+18z(ii). Using the result of (a)A(i),factorize 2(x+1)^3-12(x+1)^2+18(x+1).(b). The base of a cuboid is a square with sides of (x-2) cm each.Given that the volume of the cuboid is〔2(x+1)^3-12(x+1)^2+18(x+1)〕cm^3,using... 顯示更多 1.Factorize 2x^2(y+1)^2-4xy^2(y+1) 2.Factorize p^2 r+pr-q^2 r-qr 3a.(i)Factorize 2z^3-12z^2+18z (ii). Using the result of (a)A(i),factorize 2(x+1)^3-12(x+1)^2+18(x+1). (b). The base of a cuboid is a square with sides of (x-2) cm each.Given that the volume of the cuboid is〔2(x+1)^3-12(x+1)^2+18(x+1)〕cm^3,using the results of (a),express the height of the cuboid in terms of x. (c). If the side of a square is equal to the height of the cuboid in (b),the area of the square is (Ax^2+Bx+C)cm^3.Find the values of A,B and C.
最佳解答:
1. 2x2(y+1)2 - 4xy2(y+1) = 2x(y+1)[x(y+1) - 2y2] ... ([2x(y+1)] is the common factor of the two terms) = 2x(y+1)(xy + x - 2y2) 2. p2r + pr - q2r - qr = r(p2+p-q2-q) ... (r is the common factor of the four terms) = r(p2-q2+p-q) = r[(p+q)(p-q)+p-q] ... (a2-b2 = (a+b)(a-b) = r(p-q)(p+q+1) 3a.(i) 2z3 - 12z2 + 18z = 2z (z2 - 6z + 9) = 2z [z2 - 2(3)(z) + 32] = 2z (z-3)2 ... (a2-2ab+b2 = (a-b)2) 3.(ii) Let z be (x+1). so, 2(x+1)3-12(x+1)2+18(x+1) = 2z3 - 12z2 + 18z = 2z (z-3)2 (from(a(i))) = 2(x+1)(x+1-3)2 = 2(x+1)(x-2)2 (b) The height of the cuboid = the volume of that / the area of the base = [2(x+1)3-12(x+1)2+18(x+1)] / (x-2)2 = [2(x+1)(x-2)2] / (x-2)2 (from(a(ii))) = 2(x+1) cm 2009-02-03 16:05:33 補充： (c) because, the height of the cuboid = one side of the square so, the area of the square =[2(x+1)]2 =(2x+2)2 =4x2 + 8x + 4 cm2 (Ax2+Bx+C)cm2, given in the question, is the area of the square so, A = 4, B = 8, C = 4
其他解答:
1. 2x^2(y+1)^2-4xy^2(y+1) = 2x(y+1)[x(y+1)-2y^2] 2. (p^2)r+pr-(q^2)r-qr = r(p^2+p-q^2-q) = r[p(p+1)-q(q+1)] 3a(i) 2z^3-12z^2+18z = 2z(z^2-6z+9) = 2z(z-3)^2 (ii) 2(x+1)^3-12(x+1)^2+18(x+1) [Let z = x+1] = 2(x+1)(x+1-3)^2 = 2(x+1)(x-2)^2 b) Height = Volume/Base = [2(x+1)^3-12(x+1)^2+18(x+1)]/(x-2)^2 = [2(x+1)(x-2)^2]/(x-2)^2 = 2(x+1) c) Height = Side Area of square = Height^2 = [2(x+1)]^2 = 4(x+1)^2 = 4(x^2+2x+1) = 4x^2+8x+4 So, A = 4, B = 8, and C = 4. 2009-02-03 14:17:14 補充： Some mistakes for question 2, the correct answer is as below: (p^2)r+pr-(q^2)r-qr = r(p^2+p-q^2-q) = r(p^2-q^2+p-q) = r[(p-q)(p+q)+(p-q)] = r(p-q)(p+q+1)
中2 幾條數學問題
發問:
1.Factorize 2x^2(y+1)^2-4xy^2(y+1)2.Factorize p^2 r+pr-q^2 r-qr3a.(i)Factorize 2z^3-12z^2+18z(ii). Using the result of (a)A(i),factorize 2(x+1)^3-12(x+1)^2+18(x+1).(b). The base of a cuboid is a square with sides of (x-2) cm each.Given that the volume of the cuboid is〔2(x+1)^3-12(x+1)^2+18(x+1)〕cm^3,using... 顯示更多 1.Factorize 2x^2(y+1)^2-4xy^2(y+1) 2.Factorize p^2 r+pr-q^2 r-qr 3a.(i)Factorize 2z^3-12z^2+18z (ii). Using the result of (a)A(i),factorize 2(x+1)^3-12(x+1)^2+18(x+1). (b). The base of a cuboid is a square with sides of (x-2) cm each.Given that the volume of the cuboid is〔2(x+1)^3-12(x+1)^2+18(x+1)〕cm^3,using the results of (a),express the height of the cuboid in terms of x. (c). If the side of a square is equal to the height of the cuboid in (b),the area of the square is (Ax^2+Bx+C)cm^3.Find the values of A,B and C.
最佳解答:
1. 2x2(y+1)2 - 4xy2(y+1) = 2x(y+1)[x(y+1) - 2y2] ... ([2x(y+1)] is the common factor of the two terms) = 2x(y+1)(xy + x - 2y2) 2. p2r + pr - q2r - qr = r(p2+p-q2-q) ... (r is the common factor of the four terms) = r(p2-q2+p-q) = r[(p+q)(p-q)+p-q] ... (a2-b2 = (a+b)(a-b) = r(p-q)(p+q+1) 3a.(i) 2z3 - 12z2 + 18z = 2z (z2 - 6z + 9) = 2z [z2 - 2(3)(z) + 32] = 2z (z-3)2 ... (a2-2ab+b2 = (a-b)2) 3.(ii) Let z be (x+1). so, 2(x+1)3-12(x+1)2+18(x+1) = 2z3 - 12z2 + 18z = 2z (z-3)2 (from(a(i))) = 2(x+1)(x+1-3)2 = 2(x+1)(x-2)2 (b) The height of the cuboid = the volume of that / the area of the base = [2(x+1)3-12(x+1)2+18(x+1)] / (x-2)2 = [2(x+1)(x-2)2] / (x-2)2 (from(a(ii))) = 2(x+1) cm 2009-02-03 16:05:33 補充： (c) because, the height of the cuboid = one side of the square so, the area of the square =[2(x+1)]2 =(2x+2)2 =4x2 + 8x + 4 cm2 (Ax2+Bx+C)cm2, given in the question, is the area of the square so, A = 4, B = 8, C = 4
其他解答:
1. 2x^2(y+1)^2-4xy^2(y+1) = 2x(y+1)[x(y+1)-2y^2] 2. (p^2)r+pr-(q^2)r-qr = r(p^2+p-q^2-q) = r[p(p+1)-q(q+1)] 3a(i) 2z^3-12z^2+18z = 2z(z^2-6z+9) = 2z(z-3)^2 (ii) 2(x+1)^3-12(x+1)^2+18(x+1) [Let z = x+1] = 2(x+1)(x+1-3)^2 = 2(x+1)(x-2)^2 b) Height = Volume/Base = [2(x+1)^3-12(x+1)^2+18(x+1)]/(x-2)^2 = [2(x+1)(x-2)^2]/(x-2)^2 = 2(x+1) c) Height = Side Area of square = Height^2 = [2(x+1)]^2 = 4(x+1)^2 = 4(x^2+2x+1) = 4x^2+8x+4 So, A = 4, B = 8, and C = 4. 2009-02-03 14:17:14 補充： Some mistakes for question 2, the correct answer is as below: (p^2)r+pr-(q^2)r-qr = r(p^2+p-q^2-q) = r(p^2-q^2+p-q) = r[(p-q)(p+q)+(p-q)] = r(p-q)(p+q+1)
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