![](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)
![](https://rs.olm.vn/images/avt/0.png?1311)
(-10/3)5.(-6/5)4
= -10/3 . (-10/3)4 . (-6/5)4
= -10/3 . (-10/3.(-6/5)4
= -10/3. 44
= -10/3. 256
= -2560/3
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2^{x-2}-3.2^x=-88\)
\(\Rightarrow2^x.\frac{1}{4}-3.2^x=-88\)
\(\Rightarrow2^x.\left(\frac{1}{4}-3\right)=-88\)
\(\Rightarrow2^x.\frac{-11}{4}=-88\)
\(\Rightarrow2^x=-88:\frac{-11}{4}\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
2x-2 - 3.2x = -88
2x . 2-2 - 3.2x = -88
2x . (2-2 - 3) = -88
2x . (1/4 - 3) = -88
2x . (-11/4) = -88
2x = -88 : (-11/4)
2x = (-88).(4/-11)
2x = 32
=> x = 5
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(2^{x-2^{ }}-3\cdot2^x=-88\)
\(\Leftrightarrow2^x:2^2-3\cdot2^x=-88\)
\(\Leftrightarrow2^x\cdot\frac{1}{4}-3\cdot2^x=-88\)
\(\Leftrightarrow2^x\left(\frac{1}{4}-3\right)=-88\)
\(\Leftrightarrow2^x\cdot\left(-\frac{11}{4}\right)=-88\)
\(\Leftrightarrow2^x=-88:\left(-\frac{11}{4}\right)\Rightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2x-2 - 3.2x = -88
=> 2x-2 - 3.22.2x-2 = -88
=> 2x-2 - 3.4.2x-2 = -88
=> 2x-2 - 12.2x-2 = -88
=> 2x-2.(1 - 12) = -88
=> 2x-2.(-11) = -88
=> 2x-2 = -88 : (-11)
=> 2x-2 = 8 = 23
=> x - 2 = 3
=> x = 3 + 2 = 5
Vậy x = 5
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(2^{x-2}-3\cdot2^x=-88\)
\(\Leftrightarrow2^x\cdot\dfrac{1}{4}-3\cdot2^x=-88\)
\(\Leftrightarrow2^x\cdot\dfrac{-11}{4}=-88\)
\(\Leftrightarrow2^x=32\)
hay x=5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2^{x-2}-3\cdot2^x=-88\)
\(\Leftrightarrow\frac{2^x}{2^2}-3\cdot2^x=-88\)
\(\Leftrightarrow2^x\left(-\frac{11}{4}\right)=-88\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{2^7.3^6}{2^{11}.3^5}=\frac{3}{2^4}=\frac{3}{16}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2}{7}< \frac{1}{n}< \frac{4}{7}\)
\(\Rightarrow\frac{1}{3,5}< \frac{1}{n}< \frac{1}{1,75}\)
\(\Rightarrow3,5>n>1,75\)
\(\Rightarrow\)n \(\in\){ 2 ; 3 }
\(\frac{2}{7}< \frac{1}{n}< \frac{4}{7}\)
\(\Rightarrow n=2\)
Ta có:
\(4^1=4;4^2=16;4^3=64;4^4=256;...\)
\(\Rightarrow4^{2k}\) có chữ số tận cùng là \(6\)
\(\Rightarrow4^{2k+1}\) có chữ số tận cùng là \(4\)
Vậy \(2014^{2015}\) có dạng \(4^{2k+1}\) \(\Rightarrow\) Chữ số tận cùng là \(4\)
\(2015^{2014}\) có chữ số tận cùng là \(5\)
\(\Rightarrow A=2014^{2015}-2015^{2014}=\left(...4\right)-\left(...5\right)=...9\)
Vậy \(A=2014^{2015}-2015^{2014}\) có chữ số tận cùng là \(9\)
Ta có:\(A=2014^{2015}-2015^{2014}\)
\(A=2014^{2014}\cdot2014-2015^{2014}\)
\(A=\left(2014^2\right)^{1007}\cdot2014-2015^{2014}\)
\(A=\left(...6\right)^{1007}.2014-2015^{2014}\)
\(A=\left(...6\right)\cdot2014-2015^{2014}\)
\(A=\left(...4\right)-\left(...5\right)\)
\(A=...9\)
Vậy A có chữ số tận cùng là 9