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\(A=\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+\frac{10}{18.23}+\frac{10}{23.28}\)
\(A=2\left(\frac{5}{3.8}+\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+\frac{5}{23.28}\right)\)
\(A=2\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right)\)
\(A=2\left(\frac{1}{3}-\frac{1}{28}\right)\)
\(A=2.\frac{25}{84}=\frac{25}{42}\)
\(A=\frac{10}{3\cdot8}+\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+\frac{10}{23\cdot28}\)
\(A=10\left(\frac{1}{3\cdot8}+\frac{1}{8\cdot13}+\frac{1}{13\cdot18}+\frac{1}{18\cdot23}+\frac{1}{23\cdot28}\right)\)
\(A=\frac{10}{5}\left(\frac{5}{3\cdot8}+\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+\frac{5}{23\cdot28}\right)\)
\(A=2\cdot\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+\frac{1}{23}-\frac{1}{28}\right)\)
\(A=2\cdot\left(\frac{1}{3}-\frac{1}{28}\right)\)
\(A=2\cdot\frac{25}{84}\)
\(A=\frac{25}{42}\)
\(B=\frac{10}{3\cdot8}+\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+\frac{10}{23\cdot28}\)
\(B=2\left[\frac{5}{3\cdot8}+\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+\frac{5}{23\cdot28}\right]\)
\(B=2\left[\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right]\)
\(B=2\left[\frac{1}{3}-\frac{1}{28}\right]=\frac{25}{42}\)
Nếu ai có giải dùm mình thì giải từng phần nhưng đừng chỉ ghi kết quả nhé~
a,\(\frac{2004}{10045}\)
b,\(\frac{25}{609}\)
c,\(\frac{1000}{3549}\)
d,\(\frac{25}{258}\)
a,\(\frac{x}{10}\)+\(\frac{1}{y}\)+\(\frac{3}{10}\)
b.\(\frac{1}{x}\)+\(\frac{y}{2}\)+\(\frac{5}{8}\)
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{97.100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{97}-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\frac{99}{100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{33}{100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow x=\frac{0,33\times100}{0,33}=100\)
\(\frac{4}{8.13}+\frac{4}{13.18}+\frac{4}{18.24}+...+\frac{4}{253.258}\)
\(=\frac{4}{5}\cdot\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+...+\frac{1}{253}-\frac{1}{258}\right)\)
\(=\frac{4}{5}\cdot\left(\frac{1}{8}-\frac{1}{258}\right)\)
\(=\frac{4}{5}\cdot\frac{125}{1032}\)
\(=\frac{25}{258}\)
\(\frac{4}{8.13}+\frac{4}{13.18}+\frac{4}{18.23}+...+\frac{4}{253.258}\)
\(=\frac{4}{5}\left(\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+...+\frac{5}{253.258}\right)\)
\(=\frac{4}{5}\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+...+\frac{1}{253}-\frac{1}{258}\right)\)
\(=\frac{4}{5}\left(\frac{1}{8}-\frac{1}{258}\right)\)
\(=\frac{4}{5}.\frac{125}{1032}=\frac{25}{258}\)
a)\(\left(4\frac{5}{37}-3\frac45+8\frac{15}{29}\right)-\left(3\frac{5}{57}-6\frac{14}{29}\right)\)
=\(4\frac{5}{37}-3\frac45+8\frac{15}{29}-3\frac{5}{37}+6\frac{14}{29}\)
=\(\left(4\frac{5}{37}-3\frac{5}{37}\right)+\left(8\frac{15}{29}+6\frac{14}{29}\right)-3\frac45\)
=\(\left\lbrack\left(4-3\right)+\left(\frac{5}{37}-\frac{5}{37}\right)\right\rbrack+\left\lbrack\left(8+6\right)+\left(\frac{15}{29}\right.\right.\)+\(\frac{14}{29})\) -\(\frac{19}{5}\)
=\(1+0+14+1-\frac{19}{5}\)
=\(15+1-\frac{19}{5}\)
=\(16-\frac{19}{5}\)
=\(\frac{80}{5}-\frac{19}{5}\)
=\(\frac{61}{5}\)
\(\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+...+\frac{10}{48.53}\)
\(=\frac{10}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{48}-\frac{1}{53}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{53}\right)\)
\(=2.\frac{50}{159}=\frac{100}{159}\)