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分式的一些难题
1、当a^2+b^2=3ab时,求[1+2b^2/(a^2-b^2)]*[1+2b/(a-b)]的值.
2、当x^2-5x-2008=0时,求[(x-2)^3-(x-1)^2+1]/(x-2)的值.
3、当ab=1,求1/(1+a^2)+1/(1+b^2)的值.
(分式:“/”为除号,“^2”为平方,“^3”为立方)
人气:145 ℃ 时间:2020-06-26 07:16:53
解答
1、
由a^2+b^2=3ab 得(a-b)^2=ab
原式={[a^2-b^2+2b^2]/(a+b)(a-b)}[(a-b+2b)/(a-b)]
={3ab/(a+b)(a-b)}[(a+b)/(a-b)]
=3ab/(a-b)^2
=3ab/ab
=3
2、
x^2-5x=2008
原式={(x-2+1)[x^2-4x+4-x+2+1]-(x-1)^2}/(x-2)
=(x-1)(x^2-4x+4-x+3-x+1)/(x-2)
=(x-1)(x^2-6x+8)/(x-2)
=(x-1)(x-2)(x-4)/(x-2)
=(x-1)(x-4)
=x^2-5x+4
=2008+4
=2012
3、
1/(1+a^2)+1/(1+b^2)
=ab/(ab+a^2)+ab/(ab+b^2)
=b/(a+b)+a/(a+b)
=(a+b)/(a+b)
=1
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