![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: Sửa đề: 1/3^200
1/2^300=(1/8)^100
1/3^200=(1/9)^100
mà 1/8>1/9
nên 1/2^300>1/3^200
b: 1/5^199>1/5^200=1/25^100
1/3^300=1/27^100
mà 25^100<27^100
nên 1/5^199>1/3^300
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có \(\frac{{ - 2}}{3} < 0\) và \(\frac{1}{{200}} > 0\) nên \(\frac{{ - 2}}{3}\)<\(\frac{1}{{200}}\).
b) Ta có: \(\frac{{139}}{{138}} > 1\) và \(\frac{{1375}}{{1376}} < 1\) nên \(\frac{{139}}{{138}}\) > \(\frac{{1375}}{{1376}}\).
c) Ta có: \(\frac{{ - 11}}{{33}} = \frac{{ - 1}}{3}\) và \(\frac{{25}}{{ - 76}} = \frac{{ - 25}}{{76}} > \frac{{ - 25}}{{75}} = \frac{{ - 1}}{3}\,\,\,\, \Rightarrow \frac{{25}}{{ - 76}} > \frac{{ - 11}}{33}\).
a: -2/3<0<1/200
b: 139/138>1
1375/1376<1
=>139/138>1375/1376
c: -11/33=-1/3=-25/75<-25/76
![](https://rs.olm.vn/images/avt/0.png?1311)
TC:(1/2)^300=(1/8)^100
(1/3)^200=(1/9)^100
Vì (1/8)^100>(1/9)^100 =>(1/2)^300 >(1/3)^200
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\frac{2}{{ - 5}} = \frac{{ - 16}}{{40}}\) và \(\frac{{ - 3}}{8} = \frac{{ - 15}}{{40}}\)
Do \(\frac{{ - 16}}{{40}} < \frac{{ - 15}}{{40}}\,\, \Rightarrow \,\frac{2}{{ - 5}} < \frac{{ - 3}}{8}\).
b) Ta có: \( - 0,85 = \frac{{ - 85}}{{100}} = \frac{{ - 17}}{{20}}\). Vậy \( - 0,85\)=\(\frac{{ - 17}}{{20}}\).
c) Ta có: \(\frac{{37}}{{ - 25}} = \frac{{ - 296}}{{200}}\)
Do \(\frac{{ - 137}}{{200}} > \frac{{ - 296}}{{200}}\) nên \(\frac{{ - 137}}{{200}}\) > \(\frac{{37}}{{ - 25}}\) .
d) Ta có: \( - 1\frac{3}{{10}}=\frac{-13}{10}\) ;
\(-\left( {\frac{{ - 13}}{{ - 10}}} \right) = \frac{{-13}}{{10}}\).
Vậy \(- 1\frac{3}{{10}} =-(\frac{{-13}}{{-10}})\,\).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{2}>\frac{1}{3}\\ \Rightarrow\left(\frac{1}{2}\right)^{200}>\left(\frac{1}{3}\right)^{200}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(2^{300}>3^{200}\)
b,\(-\frac{22}{35}>-\frac{103}{137}\)
đề đúng của câu b là:
\(\frac{-22}{35}\)và \(\frac{-103}{177}\)
bn chỉ cho mình cách làm lun nha
Ta có :
\(\frac{-1}{2}^{300}=\left[\left(-\frac{1}{2}\right)^3\right]^{100}=\left(-\frac{1}{8}\right)^{100}\)
\(\frac{-1}{3}^{200}=\left[\left(-\frac{1}{3}\right)^2\right]^{100}=\frac{1}{9}^{100}\)
vì \(\left(-\frac{1}{8}\right)^{100}=\frac{1}{8}^{100}\)mà 8100 < 9100 nên \(\frac{1}{8}^{100}>\frac{1}{9}^{100}\)hay \(\left(-\frac{1}{8}\right)^{100}>\left(\frac{1}{9}\right)^{100}\)
Vậy \(\left(-\frac{1}{2}\right)^{300}>\left(-\frac{1}{3}\right)^{200}\)
\(\left(\frac{-1}{2}\right)^{300}=\left[\left(\frac{-1}{2}\right)^3\right]^{100}=\left(\frac{-1}{8}\right)^{100}\)
\(\left(\frac{-1}{3}\right)^{200}=\left[\left(\frac{-1}{3}\right)^2\right]^{100}=\left(\frac{1}{9}\right)^{100}\)
vì \(\left(\frac{-1}{8}\right)^{100}< \left(\frac{1}{9}\right)^{100}\)nên \(\left(\frac{-1}{2}\right)^{300}< \left(\frac{-1}{3}\right)^{200}\)