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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta thấy :
\(\frac{20}{39}>\frac{14}{39}\)
\(\frac{22}{27}>\frac{22}{29}\)
\(\frac{18}{43}< \frac{18}{41}\)
\(\Rightarrow\frac{20}{39}+\frac{22}{27}+\frac{18}{43}>\frac{14}{39}+\frac{22}{29}+\frac{18}{41}\)
=> A > B
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\frac{18}{-39}=-\frac{18}{39}\)
Vì -17>-18 nên \(-\frac{17}{39}>-\frac{18}{39}\)(1)
Vì 39<41 nên \(-\frac{17}{39}< -\frac{17}{41}\)(2)
Từ (1);(2)=>\(\frac{-18}{39}< -\frac{17}{39}< -\frac{17}{41}\)
b)Ta có: \(\frac{42}{-37}=-\frac{42}{37}>\frac{-42}{35}=\frac{-6}{5}=-1,2\); \(-\frac{56}{43}< -\frac{55}{43}< -\frac{55}{44}=-\frac{5}{4}=-1,25\)
Vì -1,2>-1,25 nên 42/-37>-56/43
c)Ta có:25049<25259 hay 37*677<67*377 nên 37/67<377/677
d)Ta có:\(\frac{5}{8}=\frac{10}{16}< \frac{34}{16}=\frac{17}{8}< \frac{17}{19}\); \(\frac{17}{19}< 1;\frac{22}{17}>1=>\frac{22}{17}>\frac{17}{19}\)
=>\(\frac{22}{17}>\frac{17}{19}>\frac{5}{8}\)
làm hơi lâu
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{17}{30}>\frac{51}{92}\)
b) \(\frac{-45}{47}>\frac{31}{-30}\)
c) \(\frac{22}{67}< \frac{51}{152}\)
d) \(-\frac{17}{39}< -\frac{17}{41};\frac{18}{-39}< -\frac{17}{41}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có:
A= \(\dfrac{3}{8^3}+\dfrac{7}{8^4}=\dfrac{3}{8^3}+\dfrac{3}{8^4}+\dfrac{4}{8^4}\)
B= \(\dfrac{7}{8^3}+\dfrac{3}{8^4}=\dfrac{3}{8^3}+\dfrac{4}{8^3}+\dfrac{3}{8^4}\)
Vì \(\dfrac{4}{8^4}< \dfrac{4}{8^3}\) nên A < B.
b, Ta có:
\(\dfrac{20}{39}>\dfrac{14}{39}\)
\(\dfrac{22}{27}>\dfrac{22}{29}\)
\(\dfrac{18}{43}< \dfrac{18}{41}\)
\(\Rightarrow\)\(\dfrac{20}{39}+\dfrac{22}{27}+\dfrac{18}{43}>\dfrac{14}{39}+\dfrac{22}{29}+\dfrac{18}{41}\)
Hay A > B
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\)
\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
Vi n-1<n nen n-1/n+3<n/n+3
A<B
Vì n-1/n+3<n/n+3