先化简,再求值;[(2a-b/a+b)-(b/a-b)]÷{(a-2b/a+b)+[ab/(a-b)²]},其中b/a=-1/2
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人气:352 ℃ 时间:2019-09-01 08:34:42
解答
先化简,再求值;[(2a-b/a+b)-(b/a-b)]÷{(a-2b/a+b)+[ab/(a-b)²]},其中b/a=-1/2
解,得:
[(2a-b)(a-b)-b(a+b)]/[(a+b)(a-b)]÷[(a-2b)(a-b)^2+ab(a+b)]/[(a+b)(a-b)^2]
==[(2a-b)(a-b)-b(a+b)]/[(a+b)(a-b)]* [(a+b)(a-b)^2]/[(a-2b)(a-b)^2+ab(a+b)]
==[(2a-b)(a-b)-b(a+b)](a-b)/ [(a-2b)(a-b)^2+ab(a+b)]
==2(a^3-3a^2b+2ab^2)/(a^3-3a^2b+7ab^2-2b^3)
因为b/a==-1/2
b==-a/2
把b带入2(a^3-3a^2b+2ab^2)/(a^3-3a^2b+7ab^2-2b^3),得
(2a^3+3a^3+a^3)/[(4a^3+6a^3+7a^3+a^3)/4]
==6a^3/([18a^3)/4]
==6a^3*(4/18a^3)
==4/3
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