初二代数计算1)(b-c)/(a²-ab-ac+bc)-(c-a)/(b²-bc-ab+ac)+(a-b)/(c²-ac-bc+ab)(2)1/a+1/b(1+1/a)+1/c(1+1/a)(1+1/b)+1/d(1+1/a)(1+1/b)(1+1/c)-(1+1/a)(1+1/b)(1+1/c)(1+1/d)
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初二代数计算1)(b-c)/(a²-ab-ac+bc)-(c-a)/(b²-bc-ab+ac)+(a-b)/(c²-ac-bc+ab)(2)1/a+1/b(1+1/a)+1/c(1+1/a)(1+1/b)+1/d(1+1/a)(1+1/b)(1+1/c)-(1+1/a)(1+1/b)(1+1/c)(1+1/d)
初二代数计算
1)(b-c)/(a²-ab-ac+bc)-(c-a)/(b²-bc-ab+ac)+(a-b)/(c²-ac-bc+ab)
(2)1/a+1/b(1+1/a)+1/c(1+1/a)(1+1/b)+1/d(1+1/a)(1+1/b)(1+1/c)-(1+1/a)(1+1/b)(1+1/c)(1+1/d)
初二代数计算1)(b-c)/(a²-ab-ac+bc)-(c-a)/(b²-bc-ab+ac)+(a-b)/(c²-ac-bc+ab)(2)1/a+1/b(1+1/a)+1/c(1+1/a)(1+1/b)+1/d(1+1/a)(1+1/b)(1+1/c)-(1+1/a)(1+1/b)(1+1/c)(1+1/d)
解,1,(b-c)/(a²-ab-ac+bc)-(c-a)/(b²-bc-ab+ac)+(a-b)/(c²-ac-bc+ab)
=(b-c)/(a-b)(a-c)-(c-a)/(b-a)(b-c)+(a-b)/(c-a)(c-b) 令a-b=x,b-c=y,则a-c=x+y
原式=[y/x(x+y)]-[(x+y)/xy]+[x/(x+y)y]=-2/(x+y)=-2/(a-c)
2,原式=1/a+1/b(1+1/a)+1/c(1+1/a)(1+1/b)+(1/d)-1-1/d) (1+1/a)(1+1/b)(1+1/c)
=1/a+1/b(1+1/a)+(1/c-1-1/c)(1+1/a)(1+1/b)
=1/a+(1/b-1-1/b)(1+1/a)
=1/a-1-1/a
=-1
1)分母为(a-b)(b-c)(c-a)
你自己做吧,给你思路了
2)跟一类似的