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\(\frac{3-\frac{3}{7}+\frac{3}{13}-\frac{3}{2018}}{7-\frac{7}{20}+\frac{7}{13}-\frac{7}{2018}}\)
\(=\frac{3\left(1-\frac{1}{20}+\frac{1}{13}-\frac{1}{2018}\right)}{7\left(1-\frac{1}{20}+\frac{1}{13}-\frac{1}{2018}\right)}\)
\(=\frac{3}{7}\)
A) \(\frac{1}{2}\cdot\left(\frac{2}{9}+\frac{3}{7}-\frac{5}{27}\right)\)
\(=\frac{1}{2}\cdot\frac{1}{2}\)
\(=\frac{1}{4}\)
B) \(\left(\frac{-5}{28}+1.75+\frac{8}{35}\right):\left(-3\frac{9}{20}\right)\)
\(=\left(\frac{-5}{28}+\frac{7}{4}+\frac{8}{35}\right):\frac{-69}{20}\)
\(=\frac{14}{5}:\frac{-69}{20}\)
\(=\frac{-56}{69}\)
Câu hỏi của Lê Tiến Cường - Toán lớp 6 - Học toán với OnlineMath
\(A=\frac{3}{2}+\frac{7}{6}+\frac{13}{12}+...+\frac{10101}{10100}=\frac{2+1}{2}+\frac{6+1}{6}+\frac{12+1}{12}+...+\frac{10100+1}{10100}\)
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{6}\right)+\left(1+\frac{1}{12}\right)+....+\left(1+\frac{1}{10100}\right)\)
\(A=\left(1+\frac{1}{1\times2}\right)+\left(1+\frac{1}{2\times3}\right)+\left(1+\frac{1}{3\times4}\right)+...+\left(1+\frac{1}{100\times101}\right)\)
\(A=\left(1+1+1+....+1\right)+\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{100\times101}\right)\)
\(A=100+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{100}-\frac{1}{101}\right)\)
\(A=100+1-\frac{1}{101}=101-\frac{1}{101}< 101=B\)
\(\Rightarrow A< B\)
So easy
\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}+\frac{4}{14}-\frac{2}{13}}\times\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{64}-\frac{3}{256}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{\frac{2}{6}+\frac{2}{14}-\frac{2}{26}}{\frac{4}{6}+\frac{4}{14}-\frac{4}{26}}\times\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{64}-\frac{3}{356}}{\frac{4}{4}-\frac{4}{16}+\frac{4}{64}-\frac{4}{256}}+\frac{5}{8}\)
\(=\frac{2\left(\frac{1}{6}+\frac{1}{14}-\frac{1}{26}\right)}{4\left(\frac{1}{6}+\frac{1}{14}-\frac{1}{26}\right)}\times\frac{3\left(\frac{1}{4}-\frac{1}{16}+\frac{1}{64}-\frac{1}{356}\right)}{4\left(\frac{1}{4}-\frac{1}{16}+\frac{1}{64}-\frac{1}{256}\right)}+\frac{5}{8}\)
\(=\frac{2}{4}\times\frac{3}{4}+\frac{5}{8}\)
\(=\frac{1}{2}\times\frac{3}{4}+\frac{5}{8}\)
\(=\frac{3}{8}+\frac{5}{8}\)
\(=\frac{8}{8}=1\)
\(\frac{\frac{109}{3.7.13}}{\frac{361}{3.14.13}}\)\(\frac{\frac{153}{256}}{\frac{51}{64}}\)+5/8
=\(\frac{327}{722}\)+5/8
=\(\frac{3113}{2888}\)
\(\left(\frac{3}{4}-\frac{13}{11}+\frac{7}{5}\right)-\left(3-\frac{1}{2}-\frac{35}{11}\right)+\left(\frac{11}{4}-\frac{2}{5}\right)\)
= \(\frac{3}{4}-\frac{13}{11}+\frac{7}{5}-3+\frac{1}{2}+\frac{35}{11}+\frac{11}{4}-\frac{2}{5}\)
= \(\left(\frac{3}{4}+\frac{11}{4}+\frac{1}{2}\right)\left(-\frac{13}{11}+\frac{35}{11}\right)+\left(\frac{7}{5}-\frac{2}{5}\right)-3\)
= \(8+2+1-3\)
= \(8\)
#)Giải :
\(\left(\frac{3}{4}-\frac{13}{11}+\frac{7}{5}\right)-\left(3-\frac{1}{2}-\frac{35}{11}\right)+\left(\frac{11}{4}-\frac{2}{5}\right)\)
\(=\frac{3}{4}-\frac{13}{11}+\frac{7}{5}-3+\frac{1}{2}+\frac{35}{11}+\frac{11}{4}-\frac{2}{5}\)
\(=\left(\frac{3}{4}+\frac{11}{4}\right)+\left(-\frac{13}{11}+\frac{35}{11}\right)+\left(\frac{7}{5}-\frac{2}{5}\right)-3+\frac{1}{2}\)
\(=\frac{7}{2}+2+1-3+\frac{1}{2}\)
\(=\frac{7}{2}+\frac{1}{2}\)
\(=4\)
A=\(11\frac{3}{13}-\left(2\frac{4}{7}+5\frac{3}{13}\right)\)
=\(11\frac{3}{13}-5\frac{3}{13}-2\frac{4}{7}\)
=\(6-\frac{18}{7}\)
\(\frac{521}{87}\)
k cho a nha
A=\(11\frac{3}{13}-2\frac{4}{7}-5\frac{3}{13}\)
A=\(11\frac{3}{13}-5\frac{3}{13}-2\frac{4}{7}\)
A=\(6-\frac{18}{7}\)
A=\(\frac{24}{7}\)
\(\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}\)
\(=\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}\)
\(=\left(1+\frac{1}{2}\right)-\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)-\left(\frac{1}{4}+\frac{1}{5}\right)+\left(\frac{1}{5}+\frac{1}{6}\right)-\left(\frac{1}{6}+\frac{1}{7}\right)\)
\(+\left(\frac{1}{7}+\frac{1}{8}\right)-\left(\frac{1}{8}+\frac{1}{9}\right)\)
\(=1+\frac{1}{2}-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-\frac{1}{4}-\frac{1}{5}+\frac{1}{5}+\frac{1}{6}-\frac{1}{6}-\frac{1}{7}+\frac{1}{7}+\frac{1}{8}-\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}=\frac{8}{9}\)
\(\frac{3}{5}-\frac{-7}{10}+\frac{13}{20}=\frac{12}{20}-\frac{-14}{20}+\frac{13}{20}=\frac{12-\left(-14\right)+13}{20}=\frac{12+14+13}{20}=\frac{39}{20}< \frac{40}{20}=2\)
Vậy \(\frac{3}{5}-\frac{-7}{10}+\frac{13}{20}< 2\)