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\(Z=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\)
\(Z=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{80.81}\)
\(Z=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{80}-\frac{1}{81}\)
\(Z=\frac{1}{1}-\frac{1}{81}=\frac{80}{81}\)
\(Z=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{6480}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{80.81}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+.....+\frac{1}{80}-\frac{1}{81}\)
\(=1-\frac{1}{81}\)
\(=\frac{80}{81}\)
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Bài 1:
a) Ta có: \(\frac{3}{8}+\frac{-5}{6}\)
\(=\frac{3}{8}-\frac{5}{6}\)
\(=\frac{9}{24}-\frac{20}{24}\)
\(=-\frac{11}{24}\)
b) Ta có: \(\frac{15}{12}-\frac{-1}{4}\)
\(=\frac{15}{12}+\frac{1}{4}\)
\(=\frac{15}{12}+\frac{3}{12}\)
\(=\frac{18}{12}=\frac{3}{2}\)
Bài 2:
a) Ta có: \(-\frac{1}{12}-\left(2\frac{5}{8}-\frac{1}{3}\right)\)
\(=-\frac{1}{12}-\frac{21}{8}+\frac{1}{3}\)
\(=\frac{-2}{24}-\frac{63}{24}+\frac{8}{24}\)
\(=\frac{-57}{24}\)
\(=-\frac{19}{8}\)
b) Ta có: \(\frac{-5}{6}-\left(\frac{-3}{8}+\frac{1}{10}\right)\)
\(=\frac{-5}{6}+\frac{3}{8}-\frac{1}{10}\)
\(=\frac{-100}{120}+\frac{45}{120}-\frac{12}{120}\)
\(=\frac{-67}{120}\)
c) Ta có: \(-1.75-\left(\frac{-1}{9}-2\frac{1}{18}\right)\)
\(=-\frac{7}{4}+\frac{1}{9}+\frac{37}{18}\)
\(=\frac{-63}{36}+\frac{4}{36}+\frac{74}{36}\)
\(=\frac{15}{36}=\frac{5}{12}\)
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\(A=\frac{-5}{12}+\frac{4}{37}+\frac{17}{12}-\frac{41}{37}=(\frac{-5}{12}+\frac{17}{12})+(\frac{4}{37}-\frac{41}{37})=\frac{12}{12}+\frac{-37}{37}=1+(-1)=0\)
\(B=\frac{1}{2}-\frac{43}{101}+\frac{-1}{3}-\frac{1}{6}=\frac{-43}{101}+(\frac{1}{2}+\frac{-1}{3}-\frac{1}{6})=\frac{-43}{101}+(\frac{3}{6}+\frac{-2}{6}-\frac{1}{6})=\frac{-43}{101}+0=\frac{-43}{101}\)
\(A=\frac{-5}{12}+\frac{4}{37}+\frac{17}{12}-\frac{41}{37}.\)
\(A=\left(\frac{-5}{12}+\frac{17}{12}\right)-\left(\frac{41}{37}-\frac{4}{37}\right)\)
\(A=1-1=0\)
\(B=\frac{1}{2}-\frac{43}{101}+\left(\frac{-1}{3}\right)-\frac{1}{6}\)
\(B=\left(\frac{1}{2}+\left(\frac{-1}{3}\right)-\frac{1}{6}\right)-\frac{43}{101}\)
\(A=0-\frac{43}{101}=\frac{-43}{101}\)
\(C=\frac{-5}{6}\cdot\frac{12}{-7}\cdot-\frac{21}{15}\)
\(C=\frac{-5}{2.3}\cdot\frac{3.2.2}{-7}\cdot\frac{3.\left(-7\right)}{3.5}\)
\(C=\frac{-2}{1}=-2\)
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\(D=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
Làm tắt nha :
\(D=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(D=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(D=\frac{1}{2}.\frac{98}{99}-\frac{1}{2}.\frac{98}{100}\)
\(D=\frac{1}{2}\left(\frac{98}{99}-\frac{98}{100}\right)\)
Tự tính nốt nha
A = 1/2 + 1/6 + 1/12 + .. + 1/6480
A= 1/1 x 2 + 1/2 x 3 + 1/3 x 4 + ... + 1/80 x 81
A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/80 - 1/81
A = 1 - 1/81
A = 80/81
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\)
\(A=\frac{1}{1}\times2+\frac{1}{2}\times3+\frac{1}{3}\times4+...+\frac{1}{80}\times81\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{80}-\frac{1}{81}\)
\(A=1-\frac{1}{81}\)
\(A=\frac{80}{81}\)