已知ab≠1,且满足2a²+2008a+3=0和3b²+2008b+2=0,则()
A.3a-2b=0
B.2a-3b=0
C.3a+2b=0
D.2a+3b=0
人气:369 ℃ 时间:2020-05-05 10:05:02
解答
3b²+2008b+2=0可得21/b²+20081/b+3=0
a、1/b是2x²+2008s+3=0两个根.两根之积=3/2
所以选B唉 我还是太笨了 没看明白a、1/b是2x²+2008x+3=0两个根,由根与系数关系得a*1/b=3/2
即2a=3b也就是选B
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