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a) \(\frac{4}{11}-\frac{7}{15}+\frac{7}{11}-\frac{5}{15}\)
\(=\left(\frac{4}{11}+\frac{7}{11}\right)-\left(\frac{7}{15}+\frac{5}{15}\right)\)
\(=1-\frac{4}{5}\)
\(=\frac{1}{5}\)
b) \(\frac{7}{3}-\frac{4}{9}-\frac{1}{3}-\frac{5}{9}\)
\(=\left(\frac{7}{3}-\frac{1}{3}\right)-\left(\frac{4}{9}+\frac{5}{9}\right)\)
\(=2-1\)
\(=1\)
c) \(\frac{1}{4}+\frac{7}{33}-\frac{5}{3}\)
\(=\frac{-1}{4}+\frac{-16}{11}\)
\(=\frac{-75}{44}\)
d) \(\frac{-3}{4}\times\frac{8}{11}-\frac{3}{11}\times\frac{1}{2}\)
\(=\frac{-6}{11}-\frac{3}{22}\)
\(=\frac{15}{22}\)
e) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+\frac{1}{11\times13}+\frac{1}{13\times15}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(=\frac{1}{3}-\frac{1}{15}\)
\(=\frac{4}{15}\)
\(a,\)\(\frac{1}{2}.\left(\frac{2}{9}+\frac{3}{7}-\frac{5}{37}\right)\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2.259}{2331}+\frac{3.333}{2331}-\frac{5.63}{2331}\right)\)
\(\Rightarrow\frac{1}{2}.\left(\frac{518}{2331}+\frac{999}{2331}-\frac{315}{2331}\right)\)
\(\Rightarrow\frac{1}{2}.\left(\frac{518+999-315}{2331}\right)\)
\(\Rightarrow\frac{1}{2}.\frac{1202}{2331}\)
\(\Rightarrow\frac{601}{2331}\)
\(b,\)\(\left(-\frac{5}{28}+1,75+\frac{8}{35}\right):\left(-3\frac{9}{20}\right)\)
\(\Rightarrow\frac{9}{5}:\left(-\frac{69}{20}\right)\)
\(=-\frac{12}{23}\)
\(c,\)\(\frac{1}{3}.\frac{1}{5}-\frac{7}{27}.\frac{36}{14}\)
\(=\frac{1}{15}-\frac{2}{3}\)
\(=\frac{1}{15}-\frac{10}{15}=-\frac{9}{15}=-\frac{3}{5}\)
\(d,\)\(70,5-528:\frac{15}{2}\)
\(=70,5-\frac{352}{5}\)
\(=\frac{1}{10}\)
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a)\(=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}\)
\(=\left(\frac{-3}{7}+\frac{3}{7}\right)-\left(\frac{15}{26}+\frac{2}{13}\right)\)
\(=0-\frac{19}{26}\)
\(=-\frac{19}{26}\)
c)\(=\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)
\(=\frac{-11}{23}.2-\frac{1}{23}\)
\(=\frac{-22}{23}-\frac{1}{23}\)
\(=-1\)
a)Ta có: \(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)
\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)
... . . . .
\(\frac{3}{n\left(n+3\right)}=\frac{1}{n}-\frac{1}{n+3}\)
\(\Leftrightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+3}< 1^{\left(đpcm\right)}\)
b) Ta có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
Suy ra \(\frac{2}{5}< S\) (1)
Ta lại có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
Từ đó suy ra S < 8/9
Từ (1) và (2) suy ra đpcm
a) \(x.\left(\frac{2}{3}-\frac{3}{2}\right)=\frac{5}{12}\)
\(\Leftrightarrow x.\left(-\frac{5}{6}\right)=\frac{5}{12}\)
\(\Leftrightarrow x=-\frac{1}{2}\)
b) \(\frac{7}{9}:\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)
\(\Leftrightarrow2+\frac{3}{4}x=\frac{21}{8}\)
\(\Leftrightarrow\frac{3}{4}x=\frac{5}{8}\)
\(\Leftrightarrow x=\frac{5}{6}\)
c) \(\frac{3}{5}.\left(3x-3,7\right)=-\frac{57}{10}\)
\(\Leftrightarrow3x-3,7=-\frac{19}{2}\)
\(\Leftrightarrow3x=-\frac{29}{5}\)
\(\Leftrightarrow x=-\frac{29}{15}\)
\(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(3B-B=\left(1+\frac{1}{3}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\)
\(2B=1-\frac{1}{3^8}\)
\(B=\frac{1-\frac{1}{3^8}}{2}\)
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