分式方程简算求[(2a-b-c)(b-c)]/[(a-b)(a-c)(b-c)]-[(2b-a-c)(a-c)]/[(a
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分式方程简算
求[(2a-b-c)(b-c)]/[(a-b)(a-c)(b-c)]-[(2b-a-c)(a-c)]/[(a-b)(b-c)(a-c)]+[(2c-a-b)(a-b)]/[(b-c)(a-c)(a-b)]的简算
求[(2a-b-c)(b-c)]/[(a-b)(a-c)(b-c)]-[(2b-a-c)(a-c)]/[(a-b)(b-c)(a-c)]+[(2c-a-b)(a-b)]/[(b-c)(a-c)(a-b)]的简算
[(2a-b-c)(b-c)]/[(a-b)(a-c)(b-c)]-[(2b-a-c)(a-c)]/[(a-b)(b-c)(a-c)]+[(2c-a-b)(a-b)]/[(b-c)(a-c)(a-b)]
=[(2a-b-c)(b-c)-(2b-a-c)(a-c)+(2c-a-b)(a-b)]/[(b-c)(a-c)(a-b)]
=[(2ab-2ac-b^2+c^2)+(-2ab+2bc+a^2-c^2)+(2ac-2bc-a^2+b^2)]/[(b-c)(a-c)(a-b)]
=0/[(b-c)(a-c)(a-b)]
=0
=[(2a-b-c)(b-c)-(2b-a-c)(a-c)+(2c-a-b)(a-b)]/[(b-c)(a-c)(a-b)]
=[(2ab-2ac-b^2+c^2)+(-2ab+2bc+a^2-c^2)+(2ac-2bc-a^2+b^2)]/[(b-c)(a-c)(a-b)]
=0/[(b-c)(a-c)(a-b)]
=0
化简(2a-b-c)/(a+b)(a-c)+(2b-c-a)/(b-c)(b-a)+(2c-a-b)/(c-b)(c-a
(a+b-c)(a-b+c)
1.计算:(1)(a-b)(a-c)分之2a-b-c + (b-c)(b-a)分之2b-c-a + (c-b)(c-a)
(a+b-c)(a-b+c)-(a-b-c)(a+b+c),
已知:(a+b-c)/c=(b+c-a)/a=(c+a-b)/b,a+b+c≠0.求证::(a+b)(b+c)(c+a)
(a+b+c)^2-(a-b-c)^2
(a+2b-c)(a-2b-c)
已知a ,b ,c 为正数,求证 a^2a × b^2b × c^2c ≥a^(b+c) × b^(c+a) × c^(
初二分式的加减法2a-b-c/(a-b)(a-c)+ 2b-c-a/(b-c)(b-a)+ 2c-b-a/(c-b)(c
计算(a+b-c)(a-b-c)-(a+b-c)^2
5a-{-3b+[6c-2a-(a-c)]}-[9a-(7b+c)]
-.化简分式.[(b-c)^2]/[(a-b)(a-c)]+[(c-a)^2]/[(b-a)(b-c)]+[(a-b)^