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A=33. \(\left(1-\frac{2}{3}\right)\left(1-\frac{2}{5}\right)...\left(1-\frac{2}{99}\right)\)
A=33.\(\frac{1}{3}.\frac{3}{5}....\frac{97}{99}\)
A=33.\(\frac{1}{99}\)
A=\(\frac{33}{99}=\frac{1}{3}\)
1) ( \(\frac{55}{3}\): 15 + \(\frac{26}{3}\) . \(\frac{7}{2}\)) : [(\(\frac{37}{3}\) + \(\frac{62}{7}\)) . \(\frac{7}{18}\)] : \(\frac{-1704}{445}\)
= ( \(\frac{55}{3}\). \(\frac{1}{15}\) + \(\frac{91}{3}\)) : [ \(\frac{445}{21}\) . \(\frac{7}{18}\)] . \(\frac{-445}{1704}\)
= ( \(\frac{11}{9}\)+ \(\frac{91}{3}\)) : \(\frac{445}{54}\). \(\frac{-445}{1704}\) = \(\frac{284}{9}\). \(\frac{54}{445}\). \(\frac{-445}{1704}\)= \(\frac{284}{9}\). (\(\frac{54}{445}\). \(\frac{-445}{1704}\))
= \(\frac{284}{8}\). \(\frac{-9}{284}\)
= \(\frac{-9}{8}\)
Bài 2:
a: =>x/7=1/21
=>x=1/3
c: =>x(3x-2)=0
=>x=0 hoặc x=2/3
Bài1:
a: \(=\left(-\dfrac{7}{3}\right)^{3-2}=\dfrac{-7}{3}\)
b: \(=\left(-\dfrac{4}{9}\right)^{1-3}=\left(-\dfrac{4}{9}\right)^{-2}=\dfrac{81}{16}\)
c: \(=\left(\dfrac{1}{5}\right)^{10-7}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}\)
\(S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(2S=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(2S-S=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)+\left(1+\frac{1}{2}+...+\frac{1}{2^{10}}\right)\)
\(2S-S=S=2-\frac{1}{2^{10}}\)
\(S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(2S=2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(2S=3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
\(S=2S-S\)
\(S=3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(S=3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}-1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{10}}\)
\(S=2-\frac{1}{2^{10}}\)
\(\frac{3}{2x+1}\)+ \(\frac{10}{4x+2}\) - \(\frac{6}{6x+3}\) = \(\frac{12}{26}\)
\(\dfrac{3}{2x+1}\) + \(\dfrac{2.5}{2\left(2x+1\right)}\) - \(\dfrac{2.3}{3\left(2x+1\right)}\) = \(\dfrac{6}{13}\)
\(\dfrac{3}{2x+1}\) + \(\dfrac{5}{2x+1}\) - \(\dfrac{2}{2x+1}\) = \(\dfrac{6}{13}\)
\(\dfrac{3}{2x+1}\) + \(\dfrac{5}{2x+1}\) + \(\dfrac{-2}{2x+1}\) = \(\dfrac{6}{13}\)
\(\dfrac{6}{2x+1}\) = \(\dfrac{6}{13}\)
Ta có: \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}\)= \(\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}\)
= \(\dfrac{3+5-2}{2x+1}=\dfrac{6}{2x+1}=\dfrac{12}{26}\) \(\Rightarrow156=24x+12\Rightarrow24x=144\Rightarrow x=6\)
Vậy x=6.
Học tốt nha 
\(a)\frac{62}{7}\cdot x=\frac{29}{9}\div\frac{3}{56}\)
\(\Rightarrow\frac{62}{7}\cdot x=\frac{29}{9}\cdot\frac{56}{3}\)
\(\Rightarrow\frac{62}{7}\cdot x=\frac{1624}{27}\)
\(\Rightarrow x=\frac{1624}{27}\div\frac{62}{7}\)
\(\Rightarrow x=\frac{1624}{27}\cdot\frac{7}{62}\)
\(\Rightarrow x=\frac{11368}{1674}=\frac{5684}{837}\)
Rút gọn thử đi
\(\frac{1}{1\times2}+\frac{1}{2\times3}+.......+\frac{1}{2015\times2016}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{2015}-\frac{1}{2016}\)
\(=1-\frac{1}{2016}\)
\(=\frac{2015}{2016}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=\frac{1}{1}-\frac{1}{2016}\)
\(=\frac{2015}{2016}\)