1.利用公式(a+b)(c+d)=ac+ad+bc+bd,求2003*2006-2004^2 = ??2.若(-10-5)^2=100+25+g,則g=??3.利用平方差公式展開(-3-2x)(-3+2x)可得??4.最接近(89.7)^2的整數為??5.若403*397=a^2-b^2,則a+b=??可否給我全部的詳解...我想了解
2006-02-04 14:56:15 · 2 個解答 · 發問者 ? 4 in 科學 ➔ 數學
利用公式(a+b)(c+d)=ac+ad+bc+bd,求2003*2006-2004^2 = ??解:2003x2006-20042 = (2000+3)(2000+6)-(2000+4)2 = (20002 + 9x2000 + 18) - (20002 + 8x2000 +42) = (9x2000 - 8x2000) + (18 - 16) = 2002若(-10-5)^2=100+25+g,則g=??解:(-10-5)2 = (10+5)2 = 102 + 2x5x10 + 52 = 100 + 100 + 25 = 100 + 25 + g 所以 g = 100利用平方差公式展開(-3-2x)(-3+2x)可得??解:(-3-2x)(-3+2x) = -(2x+3)(2x-3) = -(4x2-9) = -4x2 + 9最接近(89.7)^2的整數為??解:89.72 = (90-0.3)2 = 902-2x90x0.3+0.32 = 8100-54+0.09 = 8046.09 所以最接近的整數為 8046若403*397=a^2-b^2,則a+b=??解:403x397 = (400+3)(400-3) = 4002 - 32 所以 a = 400, b = 3 故 a + b = 400 + 3 = 403
2006-02-04 15:29:13 · answer #1 · answered by Frank 7 · 0⤊ 0⤋
2.若(-10-5)^2=100+25+g,則g=??
〈-15〉^2=125+g
225=125+g
100=g
3.利用平方差公式展開(-3-2x)(-3+2x)可得??
9-6x+6x-4x^2 → 9-4x^2
4.最接近(89.7)^2的整數為??
〈90-0.3〉^2
=90^2-2.90.0.3+0.3^2
=8100-54+0.03
=8046.09
整數:8046
5.若403*397=a^2-b^2,則a+b=??
〈400+3〉〈400-3〉=a^2-b^2
=400^2-3^2
a=400 b=3
a+b=403
2006-02-04 15:19:01 · answer #2 · answered by ○惠茹● 2 · 0⤊ 0⤋