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正数a,b是方程x²-7x+10=0两实根,则lga+lgb的值为:麻烦你帮我解决一下,
人气:318 ℃ 时间:2020-05-18 12:48:03
解答
x²-7x+10=0
(x-5)(x-2) = 0
x1 =5
x2 =2
lga + lgb
= lg5 + lg2
= lg(5 × 2)
= lg10
= 1
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