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a)\(\frac{-18}{91}>\frac{-23}{114}\)
b)\(\frac{-22}{35}>\frac{-103}{117}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a=\left[\left(-\dfrac{1}{2}\right)^5\right]^{107}=\left(-\dfrac{1}{32}\right)^{107}\)
\(b=\left[\left(-\dfrac{1}{3}\right)^3\right]^{107}=\left(-\dfrac{1}{27}\right)^{107}\)
mà -1/32>-1/27
nên a>b
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a, Ta có : \(9^{2000}=\left(3^2\right)^{2000}=3^{4000}\)
Mà \(3^{4000}=3^{4000}\)
\(\Rightarrow3^{4000}=9^{2000}\)
Vậy \(3^{4000}=9^{2000}\)
b, Ta có : \(2^{332}< 2^{333}=2^{3.111}=\left(2^3\right)^{111}=8^{111}\)
\(3^{223}>3^{222}=3^{2.111}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8^{111}< 9^{111}\)
\(\Rightarrow2^{333} < 3^{222}\)
\(\Rightarrow2^{332}< 3^{223}\)
Vậy \(2^{332}< 3^{223}\)
a) \(3^{4000}\) và \(9^{2000}\)
ta có:\(9^{2000}=\left(3^2\right)^{2000}=9^{2000}\)
=>\(9^{2000}=9^{2000}\Leftrightarrow3^{4000}=9^{2000}\)
b)\(2^{332}\) và \(3^{223}\)
\(2^{332}\) <\(2^{333}\) mà \(2^{333}=\left(2^3\right)^{111}=8^{111}\)(1)
\(3^{223}\) >\(3^{222}\) mà \(3^{222}=\left(3^2\right)^{111}=9^{111}\)(2)
từ (1 và 2),suy ra:8111<9111 hay 2332<3223
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\dfrac{-119}{117}=-1-\dfrac{2}{117}\)
\(\dfrac{-117}{115}=-1-\dfrac{2}{115}\)
Vì \(\dfrac{2}{117}\) < \(\dfrac{2}{115}\) nên \(\dfrac{-119}{117}\) > \(\dfrac{-117}{115}\)
Vậy, \(\dfrac{-119}{117}\) > \(\dfrac{-117}{115}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
-22 /35 và -103/117
ta có
-22/35=-2574/4094
-103/117=-3605/4094
=>-2574/4094>-3605/4094
=>-22 /35 > -103/117
Bài làm
\(\frac{117}{212}và\frac{321}{543}\)
Vì \(\frac{117}{212}< \frac{117}{543}< \frac{321}{543}\)
\(\Rightarrow\frac{117}{212}< \frac{321}{543}\)
sai rui ban a
can tim ps trung gian khac co