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Đặt A = \(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(A=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(3A=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\)
\(3A=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\)
\(3A=\frac{1}{3}-\frac{1}{33}\)
\(3A=\frac{10}{33}\)
\(A=\frac{10}{33}:3\)
\(A=\frac{10}{99}\)
\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+....+\frac{1}{990}\)
\(=\frac{1}{3\times6}+\frac{1}{6\times9}+\frac{1}{9\times11}+....+\frac{1}{30\times30}\)
\(=\frac{1}{3}\times\left(\frac{3}{3\times6}+\frac{3}{6\times9}+....+\frac{3}{30\times33}\right)\)
\(=\frac{1}{3}\times\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+....+\frac{1}{30}-\frac{1}{33}\right)\)
\(=\frac{1}{3}\times\left(\frac{1}{3}-\frac{1}{33}\right)=\frac{1}{3}\times\frac{10}{33}=\frac{10}{99}\)
\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)=\frac{1}{3}.\frac{10}{33}=\frac{10}{99}\)
help me !! mik cần giải bài này gấp các bn giải nhanh dùm mik đi
a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{47}-\dfrac{4}{47}+\dfrac{9}{53}\right)}=\dfrac{3}{4}\)
b: \(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\cdot\dfrac{1004}{2010}=\dfrac{2008}{2010}=\dfrac{1004}{1005}\)
c: \(S=\dfrac{1}{3\cdot6}+\dfrac{1}{6\cdot9}+...+\dfrac{1}{30\cdot33}\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)
\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{3\cdot6}+\frac{1}{6\cdot9}+\frac{1}{9\cdot12}+...+\frac{1}{30\cdot33}\)
\(=\frac{1}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{30\cdot33}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(=\frac{1}{3}\cdot\frac{10}{33}=\frac{10}{99}\)
\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(=\frac{1}{3}.\frac{10}{33}\)
\(=\frac{10}{99}\)
1/1.3+1/3.5+1/5.7+.......+1/2003.2005
= 1/2.(2/1.3+2/3.5+2/5.7+.......+2/2003.2005)
= 1/2.(1 -1/3 + 1/3-1/5+1/5-1/7 + ...+ 1/2003 - 1/2005)
= 1/2.(1-1/2005)
= 1/2. 2004/2005
= 1002/2005
Ta có:
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2004}\right)\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\right)\)
\(\Rightarrow2\left(\frac{1}{1}-\frac{1}{2004}\right)=\frac{1}{2}.\frac{2003}{2004}=\frac{2003}{4008}\)
đặt A với biểu thức trên
A=\(\frac{1}{3x6}\)+\(\frac{1}{6x9}\)+......+\(\frac{1}{30x33}\)
Nhân cả 2 vế với 3 ta có
A x 3 = \(\frac{3}{3x6}\)+....+\(\frac{3}{30x33}\)
A x 3 = \(\frac{1}{3}\)-\(\frac{1}{6}\)+....+\(\frac{1}{30}\)-\(\frac{1}{33}\)
A x 3 = \(\frac{1}{3}\)-\(\frac{1}{33}\)
A x 3 = \(\frac{10}{33}\)
A = \(\frac{10}{33}\):3
A= \(\frac{10}{99}\)
cho vài k đi bà con ơi