TT
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
3 tháng 10 2022
Bài 1:
a: \(2P=2^{101}-2^{100}+2^{98}-2^{97}+...+2^3-2^2\)
=>\(3P=2^{101}-2\)
hay \(P=\dfrac{2^{101}-2}{3}\)
b: \(5Q=5^{101}-5^{100}+5^{99}-5^{98}+...+5^3-5^2+5\)
=>\(6Q=5^{101}+1\)
hay \(Q=\dfrac{5^{101}+1}{6}\)
MC
3 tháng 8 2018
\(1.a)\) Ta có: \(\left\{{}\begin{matrix}64^8=\left(8^2\right)^8=8^{16}\\16^{12}=8^{12}.2^{12}=8^{12}.\left(2^3\right)^4=8^{12}.8^4=8^{16}\end{matrix}\right.\)
Có: \(8^{16}=8^{16}\Rightarrow64^8=16^{12}\)
Vậy...
\(b)\) Ta có: \(\left\{{}\begin{matrix}\left(-5\right)^{30}=\left[\left(-5\right)^3\right]^{10}=\left(-125\right)^{10}\\\left(-3\right)^{50}=\left[\left(-3\right)^5\right]^{10}=\left(-243\right)^{10}\end{matrix}\right.\)
Có: \(\left(-125\right)^{10}< \left(-243\right)^{10}\Rightarrow\left(-5\right)^{30}< \left(-3\right)^{50}\)
Vậy...
\(c)\) Ta có: \(\left\{{}\begin{matrix}2^{27}=\left(2^3\right)^9=8^9\\3^{18}=\left(3^2\right)^9=9^9\end{matrix}\right.\)
Có: \(8^9< 9^9\Rightarrow2^{27}< 3^{18}\)
Vậy...
\(d)\) Ta có: \(\left\{{}\begin{matrix}\left(\dfrac{1}{25}\right)^{10}=\left[\left(\dfrac{1}{5}\right)^2\right]^{10}=\left(\dfrac{1}{5}\right)^{20}\\\left(\dfrac{1}{125}\right)^8=\left[\left(\dfrac{1}{5}\right)^3\right]^8=\left(\dfrac{1}{5}\right)^{24}\end{matrix}\right.\)
Có: \(\left(\dfrac{1}{5}\right)^{20}< \left(\dfrac{1}{5}\right)^{24}\Rightarrow\left(\dfrac{1}{24}\right)^{10}< \left(\dfrac{1}{125}\right)^8\)
Vậy...
\(e)\)Có: \(32^9=\left(2^5\right)^9=2^{45}< 2^{52}=\left(2^4\right)^{13}=16^{13}< 18^{13}\)
\(\Rightarrow32^9< 18^{13}\)
Vậy...
HT
1
NX
21 tháng 8 2023
\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{\dfrac{19}{1}+\dfrac{18}{2}+\dfrac{17}{3}+....+\dfrac{1}{19}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{1+\left(\dfrac{18}{2}+1\right)+\left(\dfrac{17}{3}+1\right)+\left(\dfrac{1}{19}+1\right)}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{1+\dfrac{20}{2}+\dfrac{20}{3}+...+\dfrac{20}{19}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{20.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)}\)
\(=\dfrac{1}{20}\)
HH
1
VM
3 tháng 3 2020
a) \(\left(\frac{1}{16}\right)^{200}\) và \(\left(\frac{1}{2}\right)^{1000}.\)
Ta có:
\(\left(\frac{1}{16}\right)^{200}=\left[\left(\frac{1}{2}\right)^4\right]^{200}=\left(\frac{1}{2}\right)^{800}.\)
\(\left(\frac{1}{2}\right)^{1000}.\)
Vì \(800< 1000\) nên \(\left(\frac{1}{2}\right)^{800}< \left(\frac{1}{2}\right)^{1000}.\)
\(\Rightarrow\left(\frac{1}{16}\right)^{200}< \left(\frac{1}{2}\right)^{1000}.\)
Chúc bạn học tốt!