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\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B
\(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(A=-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(-A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(-2A=2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(-2A+A=\left(2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(-A=2-\frac{1}{1024}\)
\(A=\frac{1}{1024}-2\)
\(S=1+\frac{1}{1+2}+\frac{1}{1+2+3}+..+\frac{1}{1+2+3+..+2011}\)
\(S=1+\frac{1}{2.\left(2+1\right):2}+\frac{1}{3.\left(3+1\right):2}+...+\frac{1}{2011.\left(2011+1\right):2}\)
\(S=1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2011.2012}\)
\(S=1+2\left(\frac{1}{2}-\frac{1}{\cdot3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2011}-\frac{1}{2012}\right)\)
\(S=1+2\left(\frac{1}{2}-\frac{1}{2012}\right)\)
\(S=1+2.\frac{1}{2}-2.\frac{1}{2012}\)
\(S=1+1-\frac{1}{1006}\)
\(S=\frac{2011}{1006}\)
Nho 3 tick cho mk nha
1/h=1/2(1/a+1/b)=1/2a+1/2b=(a+b)/2ab
=>(a+b/)2ab-1/h=0
quy dong len ta co
(a+b)h/2abh-2ab/2abh=0=> (ah+bh-2ab)/2abh=0 =>ah+bh-2ab=0
=>ah+bh-ab-ab=0
=>a(h-b)-b(a-h)=0
=>a(h-b)=b(a-h)
=>a/b=(a-h)(h-b)
Đặt S = \(\frac{1}{2}+\frac{1}{2^5}+\frac{1}{2^9}+...+\frac{1}{2^{101}}\)
=> 24S = 16S = \(2^3+\frac{1}{2}+\frac{1}{2^5}+...+\frac{1}{2^{97}}\)
=> 16S - S = \(2^3+\frac{1}{2}+\frac{1}{2^5}+...+\frac{1}{2^{97}}-\left(\frac{1}{2}+\frac{1}{2^5}+\frac{1}{2^9}+...+\frac{1}{2^{101}}\right)\)
=> 15S = \(2^3-\frac{1}{2^{101}}\)
=> S = \(\frac{2^3-\frac{1}{2^{101}}}{15}\)
Khi đó A = \(\frac{2^3-\frac{1}{2^{101}}}{15}:\left(2^3-\frac{1}{2^{101}}\right)=\frac{1}{15}\)
Đặt S = \(\frac{1}{7^2}+\frac{1}{7^4}+\frac{1}{7^6}+...+\frac{1}{7^{100}}\)
=> 72S = 49S = \(1+\frac{1}{7^2}+\frac{1}{7^4}+...+\frac{1}{7^{98}}\)
=> 49S - S = \(\left(1+\frac{1}{7^2}+\frac{1}{7^4}+...+\frac{1}{7^{98}}\right)-\left(\frac{1}{7^2}+\frac{1}{7^4}+\frac{1}{7^6}+...+\frac{1}{7^{100}}\right)\)
=> 48S = \(1-\frac{1}{7^{100}}\)
=> \(S=\frac{1-\frac{1}{7^{100}}}{48}\)
Khi đó A = \(\left(\frac{1-\frac{1}{7^{100}}}{48}\right):\left(1-\frac{1}{7^{100}}\right)=\frac{1}{48}\)
x là số chắn
A=(-1)^n.3^n
A+3A=4A=1+(-1)^n.3^(n+1)
với x chẵn
A= [3^(x+1)+1]/4 vô nghiệm nguyên đề sai
\(\text{Công thức tổng quát: }\frac{1}{1+2+3+...+n}=\frac{2}{\left(n+1\right).n}\)
bạn thay vào òi làm tiếp ,phần tiếp theo dễ thui