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a)
\(\frac{64}{85}< \frac{64}{81}< \frac{73}{81}\)
=>\(\frac{64}{85}< \frac{73}{81}\)
b)
\(\frac{25}{26}=\frac{25.1010}{26.1010}=\frac{25250}{26260}\)
Ta có: \(1-\frac{25250}{26260}=\frac{1010}{26260}\)
\(1-\frac{25251}{26261}=\frac{1010}{26261}\)
Vì \(\frac{1010}{26260}>\frac{1010}{26261}\) nên \(\frac{25}{26}< \frac{25251}{26261}\)
a)\(\frac{64}{85}\)<\(\frac{64}{81}\)<\(\frac{73}{81}\)
b)\(\frac{25}{26}\)=\(\frac{25250}{26260}\)=\(1\)- \(\frac{1010}{26260}\)< \(1\)- \(\frac{1010}{26261}\)= \(\frac{25251}{26261}\)
Vì 21 < 43 nên \(\frac{21}{43}< 1\)
TA có\(\frac{21}{43}< \frac{21+2}{43+2}=\frac{23}{45}\)
Vậy\(\frac{21}{43}< \frac{23}{45}.\)
Cái này mìk cũng làm tương tự.....!
Ta có:
* \(21< 23\)
* \(43< 45\)
Vậy: \(\frac{21}{43}< \frac{23}{45}\)
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\(a)\) Ta có :
\(\frac{51}{85}=\frac{3}{5}\)
\(\frac{58}{145}=\frac{2}{5}\)
Vì \(\frac{3}{5}>\frac{2}{5}\) nên \(\frac{51}{85}>\frac{58}{145}\)
Vậy \(\frac{51}{85}>\frac{58}{145}\)
\(b)\) Ta có :
\(\frac{69}{-230}=\frac{-3}{10}\)
\(\frac{-39}{143}=\frac{-3}{11}\)
Vì \(\frac{-3}{10}< \frac{-3}{11}\) nên \(\frac{69}{-230}< \frac{-39}{143}\)
Vậy \(\frac{69}{-230}< \frac{-39}{143}\)
\(c)\) Ta có :
\(1+\frac{-7}{41}=\frac{34}{41}\)
\(1+\frac{13}{-47}=\frac{34}{47}\)
Vì \(\frac{34}{41}>\frac{34}{47}\) nên \(1+\frac{-7}{41}>1+\frac{13}{-47}\) hay \(\frac{-7}{41}>\frac{13}{-47}\)
Vậy \(\frac{-7}{41}>\frac{13}{-47}\)
\(d)\) Ta có :
\(1-\frac{40}{49}=\frac{9}{49}\)
\(\frac{15}{21}=\frac{5}{7}=\frac{35}{49}< \frac{40}{49}\)
Vậy \(\frac{40}{49}>\frac{15}{21}\)
a) \(\frac{{ - 3}}{8} = \frac{{ - 3.3}}{{8.3}} = \frac{{ - 9}}{{24}}\)
Vì -9 < -5 nên \(\frac{{ - 9}}{{24}} < \frac{{ - 5}}{{24}}\)
Vậy \(\frac{{ - 3}}{8} < \frac{{ - 5}}{{24}}\).
b) Cách 1: \(\frac{{ - 2}}{{ - 5}} = \frac{2}{5}; \frac{3}{{ - 5}} = \frac{-3}{{5}}\)
Vì 2 > -3 nên \(\frac{2}{5} > \frac{-3}{{5}}\)
Vậy \(\frac{{ - 2}}{{ - 5}} > \frac{3}{{ - 5}}\).
Cách 2: \(\frac{{ - 2}}{{ - 5}} = \frac{2}{5} > 0\) mà \(\frac{3}{{ - 5}} < 0\)
\(\Rightarrow\) \(\frac{{ - 2}}{{ - 5}} > \frac{3}{{ - 5}}\).
c) \(\frac{{ - 3}}{{ - 10}} = \frac{3}{{10}} = \frac{{3.2}}{{10.2}} = \frac{6}{{20}}\)
\(\frac{{ - 7}}{{ - 20}} = \frac{7}{{20}}\)
Vì 6 < 7 nên \(\frac{6}{{20}} < \frac{7}{{20}}\) nên \(\frac{{ - 3}}{{ - 10}} < \frac{{ - 7}}{{ - 20}}\).
d) \(\frac{{ - 5}}{4} = \frac{{ - 5.5}}{{4.5}} = \frac{{ - 25}}{{20}}; \frac{{ 23}}{{-20}}=\frac{{-23}}{{20}} \)
Vì -25 < -23 nên \( \frac{{ - 25}}{{20}} < \frac{{-23}}{{20}} \)
Vậy \(\frac{{ - 5}}{4} < \frac{{23}}{{ - 20}}\).
Đặt A = \(\frac{10^{20}+1}{10^{21}+1}\)
=> 10A = \(\frac{10^{21}+10}{10^{21}+1}=1+\frac{9}{10^{21}+1}\)
Đặt B = \(\frac{10^{21}+1}{10^{22}+1}\)
=> 10B = \(\frac{10^{22}+10}{10^{22}+1}=1+\frac{9}{10^{22}+1}\)
Vì \(\frac{9}{10^{21}+1}>\frac{9}{10^{22}+1}\)
=> \(1+\frac{9}{10^{21}+1}>1+\frac{9}{10^{22}+1}\)
=> 10A > 10B
=> A > B
a) Ta có : \(\frac{22}{7}=3+\frac{1}{7}=3+\frac{3}{21}< 3+\frac{3}{10}=\frac{33}{10}\)
b) Áp dụng tính chất \(\frac{a}{b}< \frac{a+m}{b+m}\)