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Câu 2:
b: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{n\left(n+1\right)}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}\)
c: \(\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{110}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-...+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=\dfrac{1}{4}-\dfrac{1}{11}=\dfrac{7}{44}\)
\(\dfrac{1}{6}x+\dfrac{1}{12}x+\dfrac{1}{20}x+...+\dfrac{1}{2450}x=1\)
\(x\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2450}\right)\)=1
\(x\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{49\times50}\right)\)=1
\(x\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\)
\(x\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=1\)
\(x\times\)\(\dfrac{12}{25}=1\)
\(\Rightarrow x=1\div\dfrac{12}{25}\)
\(x=1\times\dfrac{25}{12}=\dfrac{25}{12}\)
vậy \(x=\dfrac{25}{12}\)
vậy \(x=2\)\(x=2\)\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{3}-\dfrac{1}{9}\)\(\left(2\dfrac{2}{9}-x\right)\)=\(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{72}\)
\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{72}\)\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{8\times9}\)\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
mình làm câu a) (1/2 + 1 ) x ( 1/3 + 1 ) x ( 1/4 + 1 ) x....x (1/999 + 1 )
=>\(\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}....\dfrac{999}{998}.\dfrac{1000}{998}\)
=>\(\dfrac{3.4.5...999.1000}{2.3.4...997.998}\)
=> \(\dfrac{1000}{2}\) = 500
minh lam cau b
\(\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\cdot\cdot\cdot\cdot\dfrac{-999}{1000}\)
vi tu 1-999 co so cac so la le nen gia tri la so am
Neu tu la duong thi ta co :
\(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\cdot\cdot\cdot\cdot\dfrac{999}{1000}\\ =\dfrac{1}{1000}\)
Nhung tu so phai la am \(\Rightarrow\)\(\dfrac{1}{1000}\) se phai la\(\dfrac{-1}{1000}\)
Vay dap an la \(\dfrac{-1}{1000}\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}=\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{8.8}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}=\dfrac{7}{8}\Rightarrow A< \dfrac{7}{8}\)
Vậy A<1
chắc sách giúp đc
ai vào rùi kết bạn tui nha hoặc text tui nha