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The equations - 5x & # 178; + 1 = 6x, and (x + 1) &# 178; = 2x are reduced to general expressions
1、5x^2+6x-1=0
2、x^2+1=0
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- 19. Given that the point (1,1 / 3) is a point on the image of the function f (x) = a ^ x (a > 0 and a ≠ 1), the sum of the first n terms of the proportional sequence {an} is f (n) - C, the first term of the sequence {BN} (BN > 0) is C, and the first n terms and Sn satisfy sn-sn-1 = √ Sn + √ sn-1 (n ≥ 2) 1) Finding the general term formula of sequence {an} and {BN} 2) If the sum of the first n terms of the sequence {1 / bnbn + 1} is TN, what is the minimum positive integer n with TN > 1000 / 2009 From the meaning of the title 1）a=1/3,an=fn-c-（f（n-1）-c） =fn-f（n-1） =-How to launch 2 / 3 * (1 / 3) ^ (n-1)? ∴Tn=1/2（1-1/2n+1)=n/2n+1＞1000/2009 The solution is n > 1000 / 9 The minimum value of n is 112
- 20. Given that the point (1,1 / 3) is a point on the image of the function f (x) = a ^ x (a > 0 and a ≠ 1), the sum of the first n terms of the proportional sequence {an} is f (n) - C, and the sequence {BN} Given that the point (1,1 / 3) is a point on the function f (x) = a ^ X image, the sum of the first n terms of the proportional sequence an is f (x) - C, and the first term of the sequence BN is C, and the first n terms and Sn satisfy SN-S (n-1) = √ Sn √ s (n-1): ① the general term formula of the sequence an and BN; ② If the sum of the first n terms of the sequence {1 / BN * B (n 1)} is TN, ask TN From the meaning of the title 1）a=1/3,an=fn-c-（f（n-1）-c）=fn-f（n-1）=-2/3*（1／3） ^（n-1） The sum of the first n terms of an is (1 / 3) ^ n - 1 ∴c=1 And ∵ SN-S (n-1) = √ Sn + √ s (n-1) ∴√Sn-√Sn-1=1 ∴√Sn=n,Sn=n^2 ∴bn=Sn-Sn-1=2n-1 2) BN is substituted into 1 / bnbn BN + 1 = 1 / (2n-1) (2n + 1) = 1 / 2 (1 / 2n-1-1 / 2n + 1) ∴Tn=1/2（1-1/2n+1)=n/2n+1＞1000/2009 The solution is n > 1000 / 9 The minimum value of n is 112 ∴√Sn-√Sn-1=1 ∴√Sn=n,Sn=n^2 How is Sn = n deduced?