\(\frac{1}{a}=\frac{1}{6}+\frac{b}{3}\) => \(\frac{6}{6a}=\frac{a}{6a}+\frac{2ab}{6a}\) => 6 = a + 2ab => a.(1 + 2b) = 6 

Vì a; b nguyên 1 + 2b \(\in\)Ư(6) mà 1 + 2b lẻ nên 1 + 2b \(\in\) {-3;-1;1;3}

Vậy .....

a) Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)=> \(2.A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)

=> \(2.A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)

\(A=1-\frac{1}{2^{10}}\)=> \(1-A=1-\left(1-\frac{1}{2^{10}}\right)=\frac{1}{2^{10}}>\frac{1}{2^{11}}\)=> đpcm

b) Đặt B = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)

Vì \(\frac{1}{2^2}

\(\frac{B}{3}=\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2006}}\)

\(\frac{2B}{3}=B-\frac{B}{3}=\frac{1}{3}-\frac{1}{3^{2006}}\)

\(2B=1-\frac{1}{3^{2005}}\Rightarrow B=\frac{1}{2}-\frac{1}{2.3^{2005}}

\(S=\frac{-4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right).n}\)

\(=-\left(\frac{1}{1}-\frac{1}{5}\right)-\left(\frac{1}{5}-\frac{1}{9}\right)-\left(\frac{1}{9}-\frac{1}{13}\right)-...-\left(\frac{1}{n-4}-\frac{1}{n}\right)\)

\(=-\frac{1}{1}+\frac{1}{5}-\frac{1}{5}+\frac{1}{9}-\frac{1}{9}+\frac{1}{13}-...-\frac{1}{n-4}+\frac{1}{n}\)

\(=-\frac{1}{1}+\frac{1}{n}=\frac{1}{n}+1\)

C,O,D thẳng hàng ⇒ góc COD = 180 độ

Ta có góc COA + góc DOA = góc COD

x + 2x = 180 độ

3x = 180 độ

x = 60 độ

⇒ 2x = 120 độ

⇒ góc AOD = 120 độ

Vậy góc AOD = 120 độ

Đùa hả ? Câu hỏi toán thì OLM xóa làm gì