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Bài 1
a: 11/12=1-1/12
23/24=1-1/24
mà -1/12>-1/24
nên 11/12>23/24
b: -3/20=-9/60
-7/12=-35/60
mà -9>-35
nên -3/20>-7/12
![](https://rs.olm.vn/images/avt/0.png?1311)
em trả lời ccaua này hi vọng thầy còn nhớ em
a) -9/4<`1/3
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1/2+1/22+1/23+...+1/22020+1/22021 > B=1/3+1/4+1/5+13/60
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\left(\dfrac{1}{243}\right)^6=\left(\dfrac{1}{3}\right)^{5\cdot6}=\left(\dfrac{1}{3}\right)^{30}\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{28}>\left(\dfrac{1}{243}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{3^4}\right)^7>\left(\dfrac{1}{243}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{81}\right)^7>\left(\dfrac{1}{243}\right)^6\)
mà \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{81}\right)^7\)
nên \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{243}\right)^6\)
\(\left(\dfrac{3}{8}\right)^5\&\left(\dfrac{5}{243}\right)^3\)
\(\left(\dfrac{3}{8}\right)^5=\left(\dfrac{90}{240}\right)^5=\dfrac{90^5}{240^5}\)
\(\left(\dfrac{5}{243}\right)^3=\dfrac{5^3}{243^3}\)
\(=>\dfrac{90^5}{240^5}>\dfrac{5^3}{243^3}\)
\(=>\left(\dfrac{3}{8}\right)^5>\left(\dfrac{5}{243}\right)^3\)
ta có A= 1+3+3^2+3^3+3^4+3^5+3^6
A= 4+9+27+81+243+729
A= 1093
ta có B=3^7-1=2187-1=2186
=> A<B
Ta có : \(A=1+3+3^2+......+3^6\)
\(\Rightarrow3A=3\left(1+3^2+.....+3^6\right)\)
\(\Rightarrow3A=3+3^3+.....+3^7\)
\(\Rightarrow3A-A=3^7-1\)
\(\Rightarrow A=\frac{3^7-1}{2}\Rightarrow A< B\)