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\(a\)) Giải:
\(A=\frac{\frac{1}{3}+\frac{1}{9}-\frac{1}{27}}{\frac{5}{3}+\frac{5}{9}-\frac{5}{27}}=\frac{\frac{1}{3}+\frac{1}{9}-\frac{1}{27}}{5.\left(\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}=\frac{1}{5}\)
\(b\)) Giải:
\(B=\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}+1-\frac{7}{11}}=\frac{\left(\frac{2}{3}-\frac{1}{4}+\frac{5}{11}\right).132}{\left(\frac{5}{12}+1-\frac{7}{11}\right).132}=\frac{88-33+60}{55+132-84}=\frac{115}{103}\)
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+....+\frac{1}{6561}\)
\(\Rightarrow\)\(3A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+....+\frac{1}{2187}\)
\(\Rightarrow\)\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{2187}\right)-\left(\frac{1}{3}+\frac{1}{9}+....+\frac{1}{6561}\right)\)
\(\Rightarrow\)\(2A=1-\frac{1}{6561}=\frac{6560}{6561}\)
\(\Rightarrow\)\(A=\frac{3280}{6561}\)
a) Ta có: \(\frac{-1}{12}-\left(2\frac{5}{8}-\frac{1}{3}\right)\)
\(=-\frac{1}{12}-\frac{21}{8}+\frac{1}{3}\)
\(=\frac{-6}{72}-\frac{189}{72}+\frac{24}{72}\)
\(=-\frac{19}{8}\)
b) Ta có: \(-1,75-\left(\frac{-1}{9}-2\frac{1}{18}\right)\)
\(=\frac{-7}{4}+\frac{1}{9}+\frac{37}{18}\)
\(=\frac{-63}{36}+\frac{4}{36}+\frac{74}{36}\)
\(=\frac{5}{12}\)
c) Ta có: \(\frac{2}{5}+\frac{-4}{3}+\frac{-1}{2}\)
\(=\frac{12}{30}+\frac{-40}{30}+\frac{-15}{30}\)
\(=-\frac{43}{30}\)
d) Ta có: \(\frac{3}{12}-\left(\frac{6}{15}-\frac{3}{10}\right)\)
\(=\frac{3}{12}-\frac{6}{15}+\frac{3}{10}\)
\(=\frac{15}{60}-\frac{24}{60}+\frac{18}{60}\)
\(=\frac{3}{20}\)
e) Ta có: \(\left(8\frac{5}{11}+3\frac{5}{8}\right)-3\frac{5}{11}\)
\(=\frac{93}{11}+\frac{29}{8}-\frac{38}{11}\)
\(=5+\frac{29}{8}=\frac{40}{8}+\frac{29}{8}=\frac{69}{8}\)
f) Ta có: \(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
\(=\frac{4}{9}\cdot\left(-7\right)+\frac{59}{9}\cdot\left(-7\right)\)
\(=\left(-7\right)\cdot\left(\frac{4}{9}+\frac{59}{9}\right)=\left(-7\right)\cdot7=-49\)
g) Ta có: \(\frac{-1}{4}\cdot13\frac{9}{11}-0,25\cdot6\frac{2}{11}\)
\(=\frac{-1}{4}\cdot\frac{152}{11}+\frac{-1}{4}\cdot\frac{68}{11}\)
\(=\frac{-1}{4}\cdot\left(\frac{152}{11}+\frac{68}{11}\right)=-\frac{1}{4}\cdot20=-5\)
h) Ta có: \(5\frac{27}{5}+\frac{27}{23}+0,5-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{52}{5}+\frac{27}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{52}{5}+\frac{43}{23}+\frac{1}{2}-\frac{5}{27}\)
\(=\frac{64584}{6210}+\frac{11610}{6210}+\frac{3105}{6210}-\frac{1150}{6210}\)
\(=\frac{78149}{6210}\)
i) Ta có: \(\frac{3}{8}\cdot27\frac{1}{5}-51\frac{1}{5}\cdot\frac{3}{8}+19\)
\(=\frac{3}{8}\cdot\frac{136}{5}-\frac{3}{8}\cdot\frac{206}{5}+\frac{3}{8}\cdot\frac{152}{3}\)
\(=\frac{3}{8}\cdot\left(\frac{136}{5}-\frac{206}{5}+\frac{152}{3}\right)=\frac{3}{8}\cdot\frac{110}{3}\)
\(=\frac{55}{4}\)
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\(B=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\div\frac{4+\frac{4}{7}+\frac{4}{9}+\frac{4}{343}}{1+\frac{1}{7}+\frac{1}{9}+\frac{1}{343}}\)
\(=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\div\frac{4\left(1+\frac{1}{7}+\frac{1}{9}+\frac{1}{3}\right)}{1+\frac{1}{7}+\frac{1}{9}+\frac{1}{3}}\)
\(=\frac{1}{2}\div4=\frac{1}{8}\)
\(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(3B-B=\left(1+\frac{1}{3}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\)
\(2B=1-\frac{1}{3^8}\)
\(B=\frac{1-\frac{1}{3^8}}{2}\)
B = 1/3 + 1/9 + 1/27 + ... + 1/6561
B = 1/3^1 + 1/3^2 + 1/3^3 + ... + 1/3^8
3B = 1 + 1/3^1 + 1/3^2 + ... + 1/3^7
3B - B = ( 1 + 1/3^1 +1/3^2 + ... + 1/3^7 ) - ( 1/3^1 + 1/3^2 + 1/3^3 + .... + 1/3^8 )
2B = 1 - 1/3^8
B = 1 - 1/3^8 / 2