\(\sqrt{1+\dfrac{1}{n}+\dfrac{1}{\left(n+1\right)^2}}\\ =\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}+\dfrac{2}{n}-\dfrac{2}{n+1}-\dfrac{2}{n\left(n+1\right)}}\\ =\sqrt{\left[1+\dfrac{1}{n}-\dfrac{1}{\left(n+1\right)}\right]^2}=\left|1+\dfrac{1}{n}-\dfrac{1}{\left(n+1\right)}\right|\)

\(\Leftrightarrow P=1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{99}-\dfrac{1}{100}=98+\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{9849}{100}\)

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\(P=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)

\(\Rightarrow\dfrac{1}{2}P=\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{101}}\)

\(\Rightarrow\dfrac{1}{2}P-P=\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{101}}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{100}}\)

\(\Rightarrow-\dfrac{1}{2}P=\dfrac{1}{2^{101}}-\dfrac{1}{2^2}\)

\(\Rightarrow P=\left(\dfrac{1}{2^{101}}-\dfrac{1}{2^2}\right):\left(-\dfrac{1}{2}\right)\)

Nhận xét nè: ở mẫu số tại các phân số có tử số + mẫu số = 2012. Vì vậy mục tiêu là tạo ra con 2012 ở các phân số của mẫu số. E xử con tử số ở phân số mẫu số sao cho ra con 2012 là được =))

B = \(\dfrac{3}{5}+\dfrac{3}{5^2}+\dfrac{3}{5^3}+...+\dfrac{3}{5^{2016}}\)

=> 5B = \(3+\dfrac{3}{5}+\dfrac{3}{5^2}+...+\dfrac{3}{5^{2015}}\)

=> 4B = \(3-\dfrac{3}{5^{2016}}\)

=> B = \(\dfrac{3-\dfrac{3}{5^{2016}}}{4}\)

3/8 + 1/2 = 7/8

9/8 - 1/6 = 23/24

a: =1/2(3/4+1)=1/2x7/4=7/8

b: =9/8-1/6=27/24-4/24=23/24

a.\(\dfrac{1}{2}\times\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{1}{2}\times\left(\dfrac{3}{4}+1\right)=\dfrac{1}{2}\times\dfrac{7}{4}=\dfrac{7}{8}\)

b.\(\dfrac{3}{4}:\dfrac{2}{3}-\dfrac{1}{6}=\dfrac{3}{4}\times\dfrac{3}{2}-\dfrac{1}{6}=\dfrac{9}{8}-\dfrac{1}{6}=\dfrac{23}{24}\)

Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

=100

Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)

\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{8}{\dfrac{1}{5}}=40\)

\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)

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