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a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
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Ta có:
\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)
\(A=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
\(B=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{10}\right)=2.\frac{9}{10}\)
\(B=\frac{9}{5}\)
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\(\dfrac{7}{2}+\dfrac{2}{25}=\dfrac{7\times25}{2\times25}+\dfrac{2\times2}{25\times2}=\dfrac{175}{50}+\dfrac{4}{50}=\dfrac{179}{50}\)
\(\dfrac{23}{5}-\dfrac{11}{3}=\dfrac{23\times3}{5\times3}-\dfrac{11\times5}{3\times5}=\dfrac{69}{15}-\dfrac{55}{15}=\dfrac{14}{15}\)
\(\dfrac{3}{7}-\dfrac{1}{14}=\dfrac{3\times2}{7\times2}-\dfrac{1}{14}=\dfrac{6}{14}-\dfrac{1}{14}=\dfrac{5}{14}\)
\(\dfrac{8}{5}:\dfrac{1}{3}=\dfrac{8}{5}\times3=\dfrac{8\times3}{5}=\dfrac{24}{5}\)
\(2:\dfrac{5}{3}=2\times\dfrac{3}{5}=\dfrac{2\times3}{5}=\dfrac{6}{5}\)
\(\dfrac{4}{5}\times\dfrac{7}{15}=\dfrac{4\times7}{5\times15}=\dfrac{28}{75}\)
Lời giải:
\(\frac{7}{2}+\frac{2}{25}=\frac{7\times 25+2\times 2}{2\times 25}=\frac{179}{50}\)
\(\frac{23}{5}-\frac{11}{3}=\frac{23\times 3-5\times 11}{5\times 3}=\frac{14}{15}\)
\(\frac{3}{7}-\frac{1}{14}=\frac{6}{14}-\frac{1}{14}=\frac{6-1}{14}=\frac{5}{14}\)
\(\frac{8}{5}: \frac{1}{3}=\frac{8}{5}\times 3=\frac{24}{5}\)
$2: \frac{5}{3}=\frac{2\times 3}{5}=\frac{6}{5}$
$\frac{4}{5}\times \frac{7}{15}=\frac{28}{75}$
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a) \(\frac{1.3+3.5+5.7+7.9}{3.6+9.10+15.14+21.18}\)
= \(\frac{1.3+3.5+5.7+7.9}{1.3.2.3+3.5.2.3+5.7.2.3+7.9.2.3}\)
= \(\frac{1.3+3.5+5.7+7.9}{1.3.6+3.5.6+5.7.6+7.9.6}\)
= \(\frac{1.3+3.5+5.7+7.9}{6.\left(1.3+3.5+5.7+7.9\right)}=\frac{1}{6}\)
Dấu "." là dấu nhân cấp 2
b) \(\frac{1.2+2.3+3.4+4.5}{3.6+6.9+9.12+12.15}\)
= \(\frac{1.2+2.3+3.4+4.5}{1.2.3.3+2.3.3.3+3.4.3.3+4.5.3.3}\)
= \(\frac{1.2+2.3+3.4+4.5}{1.2.9+2.3.9+3.4.9+4.5.9}\)
= \(\frac{1.2+2.3+3.4+4.5}{9.\left(1.2+2.3+3.4+4.5\right)}=\frac{1}{9}\)
Dấu "." là dấu nhân cấp 2
c) \(\frac{0,3+\frac{3}{7}+\frac{3}{11}}{0,4+\frac{4}{7}+\frac{4}{11}}\)= \(\frac{\frac{3}{10}+\frac{3}{7}+\frac{3}{11}}{\frac{4}{10}+\frac{4}{7}+\frac{4}{11}}\)= \(\frac{3.\left(\frac{1}{10}+\frac{1}{7}+\frac{1}{11}\right)}{4.\left(\frac{1}{10}+\frac{1}{7}+\frac{1}{11}\right)}=\frac{3}{4}\)
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a) 8/15 + 1 = 23/15
b) 3/14 + 11/14 - 2/3 = 1- 2/3 = 1/3
c) 7/15 * (5/3+2/3) = 7/15 * 7/3= 49/45
d) 6/11 + 5/11 - 3/5 = 1 - 3/5 = 2/5
23/12
11/5 - 3/4 + 7/15
= 29/20 + 7/15
= 23/12