標題:
化簡、恆等式
發問:
最佳解答:
(3/4) + [x + (1/2)]2 = x2 + x + 1 是不是恆等式？ 左式 = (3/4) + [x + (1/2)]2 = (3/4) + [x2 + x + (1/4)] = (3/4) + x2 + x + (1/4) = x2 + x + 1 = 右式 上式是恆等式。 （但若題目是(3/4) + [(x + 1)/2]2 = x2 + x + 1 則不是恆等式。） ===== AB = 4cm, BC = 5cm, CD = 6cm, AD = 7cm 你想問甚麼？ ===== (3/4) + [x + (1/2)]2 = x2 + x + 1 是不是恆等式？ 答案如第一題。 ===== (2x + 3)2 = (2x)2 + 2(2x)(3) + (3)2 = 4x2 + 12x + 9 ===== (2 - 5y)2 = (2)2 - 2(2)(5y) + (5y)2 = 4 - 20y + 25y2 ===== (2a + 3b)2 = (2a)2 + 2(2a)(3b) + (3b)2 = 4a2 + 12ab + 9b2 ===== 49 - m2 = 72 - m2 = (7 + m)(7 - m) ===== 100r2 - 49s2 = (10r)2 - (7s)2 = (10r + 7s)(10r - 7s) ===== 27a2 - 3b2 (題目 2b2 應為 3b2) = 3(9a2 - b2) = 3[(3a)2 - b2] = 3(3a + b)(3a - b) ===== -18(a + 1)2 = -18(a2 + 2a + 1) = -18a2 - 32a - 32 ===== 3x2y3/21xy? = x/7y ===== (2a - 4b)/(2b - a) = -(4b - 2a)/(2b - a) = -2(2b - a)/(2b - a) = -2 ===== (3b/5a2) * (a/b)[* = 乘號] = 3ab/5a2b = 3/5a ===== (4x2/6y3) ÷ (12x/2y2) = (4x2/6y3) * (2y2/12x) = 8x2y2/72xy3 = x/9y ===== (2a - 4b)/(3a + 6b) * (10b + 5a)/(6b - 3a) = 2(a - 2b)/3(a + 2b) * 5(2b + a)/3(2b - a) = 2(a - 2b)/3(a + 2b) * 5(a + 2b)/[-3(a -2b) = 10(a - 2b)(a + 2b)/(-9)(a + 2b)(a - 2b) = -10/9
其他解答:
化簡、恆等式
發問:
此文章來自奇摩知識+如有不便請留言告知
全部都要連式 3/4+(x+1/2)^2=x^2+x+1是不是恆等式 AB=4cm,BC=5cm,CD=6cm,AD=7cm 3/4+(x+1/2)^2=x^2+x+1是不是恆等式 (2x+3)^2 (2-5y)^2 (2a+3b)^2 49-m^2 100r^2-49s^2 27a^2-2b^2 -18(a+1)^2 化簡 3x^2y^3/ 21xy^4 2a-4b/2b-a 3b/5a^2城a/b [城=x] 4x^2/6y^3÷12x/2y^2 2a-4b/3a+6b城10b+5a/6b-3a最佳解答:
(3/4) + [x + (1/2)]2 = x2 + x + 1 是不是恆等式？ 左式 = (3/4) + [x + (1/2)]2 = (3/4) + [x2 + x + (1/4)] = (3/4) + x2 + x + (1/4) = x2 + x + 1 = 右式 上式是恆等式。 （但若題目是(3/4) + [(x + 1)/2]2 = x2 + x + 1 則不是恆等式。） ===== AB = 4cm, BC = 5cm, CD = 6cm, AD = 7cm 你想問甚麼？ ===== (3/4) + [x + (1/2)]2 = x2 + x + 1 是不是恆等式？ 答案如第一題。 ===== (2x + 3)2 = (2x)2 + 2(2x)(3) + (3)2 = 4x2 + 12x + 9 ===== (2 - 5y)2 = (2)2 - 2(2)(5y) + (5y)2 = 4 - 20y + 25y2 ===== (2a + 3b)2 = (2a)2 + 2(2a)(3b) + (3b)2 = 4a2 + 12ab + 9b2 ===== 49 - m2 = 72 - m2 = (7 + m)(7 - m) ===== 100r2 - 49s2 = (10r)2 - (7s)2 = (10r + 7s)(10r - 7s) ===== 27a2 - 3b2 (題目 2b2 應為 3b2) = 3(9a2 - b2) = 3[(3a)2 - b2] = 3(3a + b)(3a - b) ===== -18(a + 1)2 = -18(a2 + 2a + 1) = -18a2 - 32a - 32 ===== 3x2y3/21xy? = x/7y ===== (2a - 4b)/(2b - a) = -(4b - 2a)/(2b - a) = -2(2b - a)/(2b - a) = -2 ===== (3b/5a2) * (a/b)[* = 乘號] = 3ab/5a2b = 3/5a ===== (4x2/6y3) ÷ (12x/2y2) = (4x2/6y3) * (2y2/12x) = 8x2y2/72xy3 = x/9y ===== (2a - 4b)/(3a + 6b) * (10b + 5a)/(6b - 3a) = 2(a - 2b)/3(a + 2b) * 5(2b + a)/3(2b - a) = 2(a - 2b)/3(a + 2b) * 5(a + 2b)/[-3(a -2b) = 10(a - 2b)(a + 2b)/(-9)(a + 2b)(a - 2b) = -10/9
其他解答:
文章標籤
全站熱搜
留言列表
