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![](https://rs.olm.vn/images/avt/0.png?1311)
1/41>3/124>5/207>2/83
Ta co
3/124=30/1240
1/41=30/1230
5/207=30/1242
2/83=30/1245
k cho mik nha!
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\frac{-2525}{2929}=\left(-1\right)\frac{25}{29}\)
\(\frac{-217}{245}=\left(-1\right)\frac{217}{245}\)
Xét \(\frac{25}{29}\)và \(\frac{217}{245}\)
Có 25 . 245 = 6125
217 . 29 = 6293
Có 6293>6125
=> 25/29<217/245
=>-25/29>-217/245
![](https://rs.olm.vn/images/avt/0.png?1311)
A=\(\frac{3^{10}+1}{3^9+1}\)>1
=> A=\(\frac{3^{10}+1}{3^9+1}\)> \(\frac{3^{10}+1+2}{3^9+1+2}\)
=>A=\(\frac{3^{10}+1}{3^9+1}\)>\(\frac{3^{10}+3}{3^9+3}\)
=>A=\(\frac{3^{10}+1}{3^9+1}\)>\(\frac{3\left(3^9+1\right)}{3.\left(3^8+1\right)}\)
=>A=\(\frac{3^{10}+1}{3^9+1}\)>\(\frac{3^9+1}{3^8+1}\)=B
vậy A>B
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(A=\frac{3^{10}+1}{3^9+1}\) => \(A.\frac{1}{3}=\frac{3^{10}+1}{3^{10}+3}=\frac{\left(3^{10}+3\right)-2}{3^{10}+3}=1-\frac{2}{3^{10}+3}\)
\(B.\frac{1}{3}=\frac{3^9+1}{3^8+1}\Rightarrow B.\frac{1}{3}=\frac{3^9+1}{3^9+3}=\frac{\left(3^9+3\right)-2}{3^9+3}=1-\frac{2}{3^9+3}\)
Vì : \(\frac{2}{3^{10}+3}< \frac{2}{3^9+3}\) nên \(A>B\)
\(a,\frac{27}{82}< \frac{27}{83}=\frac{1}{3};\frac{26}{75}>\frac{25}{75}=\frac{1}{3}\)
nên\(\frac{27}{82}< \frac{26}{75}\)
\(b,\frac{49}{78}< \frac{52}{78}=\frac{2}{3};\frac{64}{95}>\frac{64}{96}=\frac{2}{3}\)
nên\(\frac{49}{78}< \frac{64}{95}\Rightarrow\frac{-49}{78}>\frac{64}{-95}\)
c, Rút gọn:\(\frac{2525}{2929}=\frac{25}{29};\frac{217}{245}=\frac{31}{35}\)
Ta có:\(1-\frac{25}{29}=\frac{4}{29};1-\frac{31}{35}=\frac{4}{35}\Rightarrow1-\frac{25}{29}>1-\frac{31}{35}\)
\(\Rightarrow\frac{25}{29}< \frac{31}{35}\)hay\(\frac{2525}{2929}< \frac{217}{245}\)
\(d,A=\frac{3^{10}+1}{3^9+1}=1+\frac{3}{3^9+1}\);\(B=\frac{3^9+1}{3^8+1}=1+\frac{3}{3^8+1}\)
Dễ dàng nhận thấy \(\frac{3}{3^9+1}< \frac{3}{3^8+1}\Rightarrow A< B\)
Xin lỗi bạn e, mk ko làm được. Chúc bạn học tốt
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