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giai câu a
a) ta có (\(\frac{2}{11.13}\)+\(\frac{2}{13.15}\)+.....+\(\frac{2}{19.21}\))*462 - x =19
(\(\frac{1}{11}\)-\(\frac{1}{13}\)+\(\frac{1}{13}\)-\(\frac{1}{15}\)+....+\(\frac{1}{19}\)-\(\frac{1}{21}\)) * 462 -x=19
(\(\frac{1}{11}\)-\(\frac{1}{21}\))*462-x=19
\(G=\frac{1}{1.2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)
\(G=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)
\(G=1-\frac{1}{16}\)
\(G=\frac{15}{16}\)
\(G=\frac{1}{1.2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)
\(G=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)
\(G=1-\frac{1}{16}\)
\(G=\frac{15}{16}\)
\(\frac{1}{1.3.7}+\frac{1}{3.7.9}+\frac{1}{7.9.13}+\frac{1}{9.13.15}+\frac{1}{13.15.19}\)
\(=\frac{1}{2}\left(\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+...+\frac{1}{13.15}-\frac{1}{15.19}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.3}-\frac{1}{15.19}\right)=\frac{47}{285}\)
\(\frac{1}{5.8}\)\(+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{98}{1545}\)
\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=3.\frac{98}{1545}\)
\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{98}{515}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{98}{515}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{98}{515}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{98}{515}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{103}\)
\(\Leftrightarrow x+3=103\)
\(\Leftrightarrow x\)\(=103-3\)
\(\Leftrightarrow x\)\(=100\)
Vậy x = 100
~~~~~~~Hok tốt~~~~~~~~
ta có \(\frac{1}{5.8}+\frac{1}{8.11}+...\frac{1}{x.\left(x+3\right)}\)\(=\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x.\left(x+3\right)}\right)\)\(=\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{98}{1545}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{98}{1545}:\frac{1}{3}=\frac{98}{515}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{98}{515}=\frac{1}{103}\)
\(\Rightarrow x+3=103\)
\(\Rightarrow x=100\)
nhớ k nha
\(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}\)
\(=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\)
\(=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\right)\)
\(=1+\frac{28}{5}\)
\(=\frac{33}{5}\)
Ta có:
a) \(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{9}{15}=1+1+\frac{9}{15}=1\frac{9}{15}\)
b)\(\frac{10}{18}+\frac{4}{9}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}\right)\)
\(=1+\frac{31}{128}=1\frac{31}{128}\)
\(\frac{1}{1.3.7}=\frac{1}{6}\left(\frac{1}{1.3}-\frac{1}{3.7}\right)\)
\(\frac{1}{3.7.9}=\frac{1}{6}\left(\frac{1}{3.7}-\frac{1}{7.9}\right)\)
....
\(\frac{1}{13.15.19}=\frac{1}{6}\left(\frac{1}{13.15}-\frac{1}{15.19}\right)\)
Cộng các vế với nhau ta được
\(\frac{1}{1.3.7}+\frac{1}{3.7.9}+...+\frac{1}{13.15.19}=\frac{1}{6}\left(\frac{1}{1.3}-\frac{1}{15.19}\right)=\frac{37}{3.15.19}\)
a) \(1284-16401:231+58\) x\(11=1284-71+638\)
\(=1851\)
b)\(\frac{7}{4}+1\frac{1}{2}\)x \(1\frac{3}{2}:\frac{1}{4}=\frac{7}{4}+15\)
\(=\frac{67}{4}\)
c)\(2\frac{1}{4}+1\frac{2}{5}+\frac{4}{20}\)\(=\frac{9}{4}+\frac{7}{5}+\frac{1}{5}\)
\(=\frac{77}{20}\)
d)\(\left(\frac{12}{5}+\frac{2}{3}-\frac{11}{4}\right):\frac{3}{4}=\frac{19}{60}:\frac{3}{4}\)
\(=\frac{19}{60}\)x \(\frac{4}{3}\)
\(=\frac{19}{45}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2014.2015}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-...+\frac{1}{2014}-\frac{1}{2015}=1-\frac{1}{2015}=\frac{2014}{2015}\)
\(B=\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{97.99}=\frac{2}{11}-\frac{2}{13}+...+\frac{2}{97}-\frac{2}{99}=\frac{2}{11}-\frac{2}{99}=\frac{16}{99}\)
C=(1x2)+(2x3)+...+(99x100)
3xC=1x2x3+2x3x3+...+3x99x100
3xC=1x2x3-0x1x2+2x3x4-1x2x3+...+99x100x101-98x99x100
3xC=99x100x101
3xC=999900
C=333300
Ta có: \(\frac{1}{10.11}+\frac{1}{11.12}+....+\frac{1}{49.50}\)
\(\Rightarrow\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{10}-\frac{1}{50}\)
\(=\frac{2}{25}\)
=> 2/25