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a) \(\dfrac{-5}{6}=\dfrac{-340}{408}\);\(\dfrac{7}{8}=\dfrac{357}{408}\);\(\dfrac{7}{24}=\dfrac{119}{408}\)
\(\dfrac{16}{17}=\dfrac{384}{408}\); \(\dfrac{-3}{4}=\dfrac{-306}{408}\); \(\dfrac{2}{3}=\dfrac{272}{408}\)
Do đó: \(\dfrac{-5}{6}< \dfrac{-3}{4}< \dfrac{7}{24}< \dfrac{2}{3}< \dfrac{7}{8}< \dfrac{16}{17}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{205}{107}\),\(\dfrac{20}{23}\),\(\dfrac{7}{10}\),\(\dfrac{214}{315}\),\(\dfrac{-5}{8}\),\(\dfrac{-16}{19}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{2727-101}{3.303+404}=\dfrac{2626}{909+404}=\dfrac{2626}{1313}=2\)
b) \(\dfrac{8.9-4.15}{12.7-180}=\dfrac{72-60}{84-180}=\dfrac{12}{-96}=\dfrac{-1}{8}\)
c) \(\dfrac{-19}{3^2.7.11}=\dfrac{-19}{9.7.11}=\dfrac{-19}{63.11}=\dfrac{-19}{693}\)
d) \(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\dfrac{2^{12}.3^{10}+120.6^9}{2^{12}.3^{12}-6^{11}}=\dfrac{2^2.6^{10}+20.6.6^9}{6^{12}-6^{11}}=\dfrac{4.6^{10}+20.6^{10}}{6^{11}\left(6-1\right)}=\dfrac{\left(4+20\right).6^{10}}{5.6^{11}}=\dfrac{24}{30}=\dfrac{4}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\dfrac{-5}{6}=\dfrac{-20}{24};\dfrac{7}{8}=\dfrac{21}{24};\dfrac{7}{24}=\dfrac{7}{24};\dfrac{-3}{4}=-\dfrac{18}{24};\dfrac{2}{3}=\dfrac{16}{24}\)
Do đó: \(\dfrac{-5}{6}< -\dfrac{3}{4}< \dfrac{7}{24}< \dfrac{2}{3}< \dfrac{7}{8}\)
\(\dfrac{7}{8}=\dfrac{119}{136};\dfrac{16}{17}=\dfrac{128}{136}\)
mà 119<128
nên 7/8<16/17
DO đó: -5/6<-3/4<7/24<2/3<7/8<16/17
b: \(\dfrac{-5}{8}=\dfrac{-95}{8\cdot19};\dfrac{-16}{19}=\dfrac{-128}{19\cdot8}\)
Do đó: -5/8>-16/19
\(\dfrac{7}{10}=0.7;\dfrac{20}{23}\simeq0.87;\dfrac{214}{315}\simeq0.68;\dfrac{205}{107}>1\)
Do đó: \(\dfrac{205}{107}>\dfrac{20}{23}>\dfrac{7}{10}>\dfrac{214}{315}>-\dfrac{5}{8}>-\dfrac{16}{19}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\dfrac{-7}{12}=\dfrac{-7\cdot5}{12\cdot5}=\dfrac{-35}{60};\dfrac{-1}{-15}=\dfrac{1}{15}=\dfrac{4}{60};\dfrac{-5}{4}=\dfrac{-5\cdot15}{4\cdot15}=-\dfrac{75}{60}\)
\(\dfrac{3}{-5}=\dfrac{-3}{5}=\dfrac{-3\cdot12}{5\cdot12}=-\dfrac{36}{60}\)
mà -75<-36<-35<4
nên \(-\dfrac{75}{60}< -\dfrac{36}{60}< -\dfrac{35}{60}< \dfrac{4}{60}\)
=>\(\dfrac{-5}{4}< \dfrac{3}{-5}< \dfrac{-7}{12}< \dfrac{-1}{-15}\)
b: \(\dfrac{-8}{25}+\dfrac{22}{23}+\dfrac{-17}{25}\)
\(=\left(-\dfrac{8}{25}-\dfrac{17}{25}\right)+\dfrac{22}{23}\)
\(=-1+\dfrac{22}{23}=-\dfrac{1}{23}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\dfrac{8\cdot9-4\cdot15}{12\cdot7-180}\)
\(=\dfrac{4\cdot18-4\cdot15}{12\cdot7-12\cdot15}\)
\(=\dfrac{4\cdot\left(18-15\right)}{12\cdot\left(7-15\right)}\)
\(=\dfrac{4\cdot3}{12\cdot\left(-8\right)}\)
\(=\dfrac{12}{12\cdot\left(-8\right)}=\dfrac{-1}{8}\)
b) Ta có: \(\dfrac{2^3\cdot3}{2^2\cdot3^2\cdot5}\)
\(=\dfrac{2^2\cdot2\cdot3}{2^2\cdot3\cdot3\cdot5}\)
\(=\dfrac{2}{3\cdot5}=\dfrac{2}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{3}{4}\)\(\dfrac{2}{3}\)\(\dfrac{1}{3}\)0\(\dfrac{-1}{4}\)\(\dfrac{-2}{5}\)
\(a,\dfrac{-5}{8};\dfrac{7}{10};\dfrac{16}{19};\dfrac{20}{23}\)
\(\Rightarrow\dfrac{20}{23}>\dfrac{16}{19}>\dfrac{7}{10}>\dfrac{-5}{8}\)
Vậy \(\dfrac{20}{23}>\dfrac{16}{19}>\dfrac{7}{10}>\dfrac{-5}{8}\)
\(b,\dfrac{5\times6+6\times7}{5\times5+20}\)và \(\dfrac{8\times9-4\times15}{12\times7-180}\)
Xét : \(\dfrac{5\times6+6\times7}{5\times5+20}\)
\(=\dfrac{\left(5+7\right)\times6}{5\times5+5\times4}\)
\(=\dfrac{12\times6}{\left(5+4\right)\times5}\)
\(=\dfrac{72}{9\times5}\)
\(=\dfrac{72}{45}\)
\(=\dfrac{8}{5}\)
Xét : \(\dfrac{8\times9-4\times15}{12\times7-180}\)
\(=\dfrac{72-60}{84-180}\)
\(=\dfrac{12}{-96}\)
\(=\dfrac{-1}{8}\)
Quy đồng \(\dfrac{8}{5}=\dfrac{8\times8}{5\times8}=\dfrac{64}{40}\\ \dfrac{-1}{8}=\dfrac{\left(-1\right)\times5}{8\times5}=\dfrac{-5}{40}\)
Vì \(\dfrac{64}{40}>\dfrac{-5}{40}\)
\(\Rightarrow\dfrac{8}{5}>\dfrac{-1}{8}\)
\(\Rightarrow\dfrac{5\times6+6\times7}{5\times5+20}>\dfrac{8\times9-4\times15}{12\times7-180}\)
Vậy \(\dfrac{5\times6+6\times7}{5\times5+20}>\dfrac{8\times9-4\times15}{12\times7-180}\)
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