Dấu < , > , = ??? ( lười :) )
\(\frac{9}{15}.................\frac{9}{17}\)\(,\)\(\frac{12}{25}...................\frac{36}{75}\)\(,\)\(\frac{17}{15}...............\frac{5}{4}\)
K
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Những câu hỏi liên quan
9 tháng 1 2020
Đặt A=\(\frac{13}{21}-\frac{15}{28}+\frac{17}{36}-...+\frac{197}{4851}-\frac{199}{4950}\)
\(\frac{A}{2}=\frac{13}{42}-\frac{15}{56}+\frac{17}{72}-...+\frac{197}{9702}-\frac{199}{4950}\)
\(=\frac{6+7}{6.7}-\frac{7+8}{7.8}+\frac{8+9}{8.9}-...+\frac{98+99}{98.99}-\frac{99+100}{99.100}\)
\(=\frac{1}{7}+\frac{1}{6}-\frac{1}{8}-\frac{1}{7}+\frac{1}{9}+\frac{1}{8}-...+\frac{1}{99}+\frac{1}{98}-\frac{1}{100}+\frac{1}{99}\)
\(=\frac{1}{6}-\frac{1}{100}=\frac{47}{300}\)
\(\Rightarrow A=\frac{47}{300}.2=\frac{47}{150}\)
\(\Rightarrow Q=\frac{85}{25}+\frac{9}{10}-\frac{11}{5}+\frac{47}{150}=\frac{181}{75}\)
Vậy Q=\(\frac{181}{75}\).
16 tháng 8 2017
Ta có : \(5\cdot\left(\frac{1}{5}+\frac{1}{17}\right)-\left(\frac{2}{5}+\frac{2}{17}+\frac{9}{15}+\frac{12}{68}\right)\)
\(=\) \(5\cdot\frac{1}{5}+5\cdot\frac{1}{17}-\left(\frac{2}{5}+\frac{2}{17}+\frac{3}{5}+\frac{3}{17}\right)\)
\(=\) \(1+\frac{5}{17}-\left[\left(\frac{2}{5}+\frac{3}{5}\right)+\left(\frac{2}{17}+\frac{3}{17}\right)\right]\)
\(=\) \(1+\frac{5}{17}-\left(1+\frac{5}{17}\right)\)
\(=\) \(1+\frac{5}{17}-1-\frac{5}{17}\)
\(=\)\(0\)
Vậy ...
Tk ủng hộ mk nha các bn ❣❣ C.ơn nhiều ^^
S
11 tháng 2 2020
\(A=\frac{88}{25}-2\left(\frac{9}{20}-\frac{11}{30}+\frac{13}{42}-.....-\frac{199}{9900}\right)\)
\(A=\frac{88}{25}-2\left(\frac{4+5}{4.5}-\frac{5+6}{5.6}+....-\frac{99+100}{99.100}\right)\)
\(A=\frac{88}{25}-2\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{5}-\frac{1}{6}+\frac{1}{6}+....-\frac{1}{99}-\frac{1}{100}\right)\)
\(A=\frac{88}{25}-2\left(\frac{1}{4}-\frac{1}{100}\right)=\frac{88}{25}-\frac{1}{2}+\frac{1}{50}=\frac{176-25+1}{50}=\frac{152}{50}=\frac{76}{25}\)
SO
31 tháng 8 2017
\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\)
\(=\frac{71}{19}+\frac{13}{17}+\frac{35}{43}+6\)
\(=\frac{1454}{323}+\frac{35}{43}+6\)
\(=5,...+6\)
\(=11,...\)
3 tháng 7 2018
\(Bai2a\)\(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\)
\(=\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}\)
\(=\sqrt{3}-2\)
\(VayA=\sqrt{3}-2\)
31 tháng 12 2016
c )
\(1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}}=1+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{3}{2}}}}=1+\frac{1}{1+\frac{1}{1+\frac{5}{3}}}=1+\frac{1}{1+\frac{1}{\frac{8}{3}}}=1+\frac{1}{\frac{11}{8}}=\frac{19}{11}\)
\(\frac{9}{15}< \frac{9}{17}\)
\(\frac{12}{25}=\frac{36}{75}\)
\(\frac{17}{15}< \frac{5}{4}\)