若实数a≠b,且a,b满足a²-8a+5=0,b²-8b+5=0则代数式(b-1)/(a-1)+(a-1)/(b-1)的值为
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若实数a≠b,且a,b满足a²-8a+5=0,b²-8b+5=0则代数式(b-1)/(a-1)+(a-1)/(b-1)的值为
若实数a≠b,且a,b满足a²-8a+5=0,b²-8b+5=0
则代数式(b-1)/(a-1)+(a-1)/(b-1)的值为
若实数a≠b,且a,b满足a²-8a+5=0,b²-8b+5=0则代数式(b-1)/(a-1)+(a-1)/(b-1)的值为
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+b²-2(a+b)+2]/[ab-(a+b)+1]
=[54-2×8+2]/[5-8+1]
=40/(-2)
=-20
a,b是方程x^2-8x+5=0的两个根
a+b=8,ab=5
(b-1)/(a-1)+(a-1)/(b-1)
=(b-1)^2+(a-1)^2/(a-1)(b-1)
=(a^2-2a+1+b^2-2b+1)/(a-1)(b-1)
=(8a-5-2a+1+8b-5-2b+1)/(a-1)(b-1)
=[6(a+b)-8]/[ab-(a+b)+1]
=(6*8-8)/(5-8+1)
=-20
a≠b,且a,b满足a²-8a+5=0,b²-8b+5=0
所以a和b是方程x²-8x+5=0的根
由韦达定理
a+b=8
ab=5
原式=[(b-1)²+(a-1)²]/(a-1)(b-1)
=(a²+b²-2a-2b+2)/(ab+a+b+1)
=[(a+b)²-2ab-2(a+b)+2]/(5+8+1)
=(64-10-16+2)/14
=20/7