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若实数a≠b,且a,b满足a 2 -8a+5=0,b 2 -8b+5=0,...

解:因为a,b满足a²-8a+5=0,b²-8b+5=0 所以a,b是方程x²-8x+5-0的两个根。 即:a+b=8 ab=5 a²+b²=(a+b)²-2ab=64-2×5=54. (b-1)/(a-1)+(a-1)/(b-1) =[(b-1)²+(a-1)²]/(a-1)(b-1) =[a²+b&...

两种解法参考:①解:由a^2-8a+5=0,b^2-8b+5=0得:(a-1)^2-6(a-1)-2=0,(b-1)^2-6(b-1)-2=0 (a-1)^2=6(a-1)+2,(b-1)^2=6(b-1)+2 即a-1和b-1是x^2-6x-2=0的两个根,所以 (a-1)+(b-1)=6 (a-1)(b-1)=-2 ∴(b-1)/(a-1)+(a-1)/(b-1)=[(b-1)^2+(a-1)^2]...

由题意得a+b=8,ab=5,原式=(b?1)2(a?1)(b?1)+(a?1)2(a?1)(b?1)=b2?2b+1+a2?2a+1(a?1)(b?1)=(a+b)2?2ab?2(a+b)+2ab?(a+b)+1=82?2×5?2×8+25?8+1=40?2=-20,故答案为-20.

∵a不等于b a,b满足a^2-8a+5=0,b^2-8b+5=0 ∴a+b=8 ab=5 ∴(b-1)/(a-1)+(a-1)/(b-1)=[(b-1)ˆ2+(b-1)ˆ2]/(a-1)(b-1) =[(a+b)ˆ2-2ab-2(a+b)+2]/[ab-(a+b)+1] =[64-10-16+2]/[5-8+1] =-25

∵ log8a+log4b^2=5,∴1/3log2a+log2b=5∴log2a+3log2b=15① ∵ log8b+log4a^2=7 ∴1/3log2b+log2a=7∴3log2a+log2b=21② ②-① 2log2a-2log2b=6∴log2a-log2b=3 ∴log2(a/b)=3 a/b=2^3=8

由已知条件可知,a、b为方程x 2 -8x+5=0的两根,此时△>0,∴a+b=8,ab=5,∴ b-1 a-1 + a-1 b-1 = a 2 + b 2 -2(a+b)+2 ab-(a+b)+1 = (a+b) 2 -2ab-2(a+b)+2 ab-(a+b)+1 =-20故选A

∵a,b满足a2-8a+5=0,b2-8b+5=0,∴a,b可看着方程x2-8x+5=0的两根,∴a+b=8,ab=5,b?1a?1+a?1b?1=(b?1)2+(a?1)2(a?1)(b?1)=(a+b)2?2ab?2(a+b)+2ab?(a+b)+1=82?2×5?2×8+25?8+1=-20.故选A.

若a、b不相等,那么a、b就是方程x²-8x+5=0的两个根。 a+b=8 ab=5 b/a+a/b=(a²+b²)/ab=[(a+b)²/ab]-2=54/5 如果ab相等,那么原式=2 请采纳,谢谢!

设a b 分别是 x²-8x+5=0的两个根 a+b=8 ab=5 b/a+a/b =(b²+a²)/ab =【(a+b)²-2ab】/ab =(8²-2x5)/5 =54/5 =10.8

实数a不等于b,且a,b满足a2-8a+5=0,b2-8b+5=0, 则a,b分别是方程x²-8x+5=0的两个实数根 a+b=8 ab=5 (b-1)/(a-1)+(a-1)/(b-1) =[(b-1)²+(a-1)²]/(a-1)(b-1) =[a²+b²-2(a+b)+2]/[ab-(a+b)+1] =[(a+b)²-2ab-2(a+b)+2]...

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