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Ta có: 8> 6 nên \(\frac{8}{6}>1\)
42 < 43 nên \(\frac{42}{43}\frac{42}{43}\)
\(\dfrac{6}{7}+\dfrac{7}{8}=\dfrac{48}{56}+\dfrac{49}{56}=\dfrac{97}{56}\)
\(\dfrac{4}{5}-\dfrac{2}{3}=\dfrac{12}{15}-\dfrac{10}{15}=\dfrac{2}{15}\)
\(\dfrac{1}{5}:\dfrac{2}{7}=\dfrac{1}{5}.\dfrac{7}{2}=\dfrac{7}{10}\)
\(\dfrac{4}{7}-\dfrac{11}{42}=\dfrac{24}{42}-\dfrac{11}{42}=\dfrac{13}{42}\)
\(\dfrac{2}{3}+\dfrac{3}{51}=\dfrac{34}{51}+\dfrac{3}{51}=\dfrac{37}{51}\)
Tính
\(\dfrac{6}{7}+\dfrac{7}{8}=\dfrac{48}{56}+\dfrac{49}{56}=\dfrac{48+49}{56}=\dfrac{97}{56}\)
\(\dfrac{4}{5}-\dfrac{2}{3}=\dfrac{12}{15}-\dfrac{10}{15}=\dfrac{12-10}{15}=\dfrac{2}{15}\)
\(\dfrac{1}{5}:\dfrac{2}{7}=\dfrac{1}{5}.\dfrac{7}{2}=\dfrac{1.7}{5.2}=\dfrac{7}{10}\)
\(\dfrac{4}{7}-\dfrac{11}{42}=\dfrac{24}{42}-\dfrac{1}{42}=\dfrac{24-1}{42}=\dfrac{23}{42}\)
\(\dfrac{2}{3}+\dfrac{3}{51}=\dfrac{34}{51}+\dfrac{3}{51}=\dfrac{34+3}{51}=\dfrac{37}{51}\)
\(\frac{9}{10}>\frac{8}{9}>\frac{7}{8}>\frac{6}{7}>\frac{5}{6}>\frac{4}{5}>\frac{3}{4}>\frac{2}{3}>\frac{1}{2}.\)
1) a) \(\frac{5454}{5757}-\frac{171717}{191919}=\frac{18\times3\times101}{19\times3\times101}-\frac{17\times10101}{19\times10101}=\frac{18}{19}-\frac{17}{19}=\frac{1}{19}\)
b) \(\frac{6}{5}\times\frac{7}{6}\times\frac{8}{7}\times....\times\frac{2021}{2020}=\frac{6\times7\times8\times...\times2021}{5\times6\times7\times...\times2020}=\frac{2021}{5}\)
2) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}=2\times\frac{1}{6}+2\times\frac{1}{12}+2\times\frac{1}{20}+...+2\times\frac{1}{90}\)
\(=2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{9\times10}\right)\)
\(=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=2\times\left(\frac{1}{2}-\frac{1}{10}\right)=2\times\frac{2}{5}=\frac{4}{5}\)
b)Vì \(a-1< a+1\)
=> \(\frac{1}{a-1}>\frac{1}{a+1}\)
ai tra loi tui se k cho
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