六一儿童节的英语单词:用简便方法计算:(a+b-c)(a-b-c)-(a-b-c)^2
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要详细的解题过程
(a+b-c)(a-b-c)-(a-b-c)^2
=(a-b-c)[(a+b-c)-(a-b-c)]
=(a-b-c)2b
(a+b-c)(a-b-c)-(a-b-c)^2
=(a+b-c)(a-b-c)-(a-b-c)(a-b-c)
=(a-b-c)[(a+b-c)-(a-b-c)]
=2b(a-b-c)
若再去括号,可得
2ab-2b^2-2bc
=(a-b-c)[a+b-c-(a-b-c)]
=(a-b-c)(2b)
=2ab-2b^2-2bc
(a+b-c)(a-b-c)-(a-b-c)^2
=(a-b-c)[(a+b-c)-(a-b-c)]
=(a-b-c)2b
(a+b-c)(a-b-c)-(a-b-c)^2
=(a-b-c)[(a+b-c)-(a-b-c)]
=(a-b-c)2b
=[(a-c)+b][(a-c)-b]-[(a-c)-b]^2
=(a-c)^2-b^2-(a-c)^2+2b(a-c)+b^2
=2ab-2bc
用简便方法计算:(a+b-c)(a-b-c)-(a-b-c)^2
|a|+|b-a|-|c-a|-|a+b|
(a-b+c)(a+b-c)
(A-B)*(B-C)*(A-C)
{(a-b)/(a+b)}+{(b-c)/(b+c)}+{(c-a)/(c+a)}+{(a-b)(b-c)(c-a)/(a+b)(b+c)(c+a)}
(a+b+c)×(a+b+c)=?
初一数学题..(a+b)(a-b)+c(a+b)为什么=(a+b)(a-b+c)
化简:(a+b)^2(b+c-a)(c+a - b)+(a - b)^2(a+b+c)(a+b - c)
答题!!!!!!!!!!!!!!!!!!!!2(a*a*a-3)(a-b+c)-3(b-a*a)(a+b-c)
(a+b)(a-b)+a(2b-c)因式分解