\(\dfrac{32.5^3}{2^6.5^2}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
KB
1
H9
HT.Phong (9A5)
CTVHS
8 tháng 8 2023
a) \(\dfrac{2^6\cdot5^5}{10^5}\)
\(=\dfrac{2^6\cdot5^5}{2^5\cdot5^5}\)
\(=2\)
b) \(\dfrac{3^6\cdot5^7}{15^6}\)
\(=\dfrac{3^6\cdot5^7}{3^6\cdot5^6}\)
\(=5\)
NT
1
21 tháng 5 2022
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{5}}=\dfrac{z}{\dfrac{1}{2}}=\dfrac{x-z}{\dfrac{2}{3}-\dfrac{1}{2}}=\dfrac{-13}{2}:\dfrac{1}{6}=-39\)
Do đó: x=-26; y=-117/5; z=-39/2
15 tháng 4 2022
\(\dfrac{2^6.18+2^7}{2^6.5^2-2^6.3}=\dfrac{2^6\left(18+2\right)}{2^6.\left(5^2-3\right)}=\dfrac{20}{5^2-3}=\dfrac{20}{25-3}=\dfrac{20}{22}=\dfrac{10}{11}\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 4 2022
\(=\dfrac{2^6.18+2^6.2}{2^6.25-2^6.3}=\dfrac{2^6\left(18+2\right)}{2^6\left(25-3\right)}=\dfrac{18+2}{25-3}=\dfrac{20}{22}=\dfrac{10}{11}\)
11 tháng 2 2018
\(5D=1+\dfrac{1}{5^2}-\dfrac{1}{5^3}+\dfrac{1}{5^4}-\dfrac{1}{5^5}+...+\dfrac{1}{6.5^{99}}\)
\(6D=\dfrac{5^{100}-1}{5^{100}}+\dfrac{1}{6.5^{100}}\)
\(D=\dfrac{\dfrac{5^{100}-1}{5^{100}}+\dfrac{1}{36.5^{100}}}{6}\)
TA
30 tháng 5 2021
a) `(-63)/72 = (-63:9)/(72:9)=(-7)/8`
b) `20/(-140) = (20:20)/(-140:20) = 1/(-7)=(-1)/7`
c) `(3.10)/(5.24) =(3.2.5)/(5.2 .3.4)=1/4`
d) `(6.5-6.2)/(6+3)=18/9=2`
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 1 2021
\(\dfrac{1}{u_n-1}=\dfrac{1}{\dfrac{2^n-5^n}{2^n+5^n}-1}=\dfrac{2^n+5^n}{-2.5^n}=-\dfrac{1}{2}\left[\left(\dfrac{2}{5}\right)^n+1\right]\)
\(\Rightarrow S_n=-\dfrac{1}{2}\left[\left(\dfrac{2}{5}\right)^1+\left(\dfrac{2}{5}\right)^2+...+\left(\dfrac{2}{5}\right)^n+n\right]\)
Lại có: \(\left(\dfrac{2}{5}\right)^1+\left(\dfrac{2}{5}\right)^2+...+\left(\dfrac{2}{5}\right)^n=\dfrac{2}{5}.\dfrac{1-\left(\dfrac{2}{5}\right)^n}{1-\dfrac{2}{5}}=\dfrac{2}{3}\left[1-\left(\dfrac{2}{5}\right)^n\right]\)
\(\Rightarrow S_n=-\dfrac{1}{2}\left[\dfrac{2}{3}-\dfrac{2}{3}\left(\dfrac{2}{5}\right)^n+n\right]=...\)
0
2
14 tháng 8 2017
ta có : \(A=6.7+6.7^2+6.7^3+...+6.7^{100}\)
\(\Rightarrow7A=7.\left(6.7+6.7^2+6.7^3+...+6.7^{100}\right)\)
\(7A=6.7^2+6.7^3+6.7^4+...+6.7^{101}\)
\(\Rightarrow7A-A=6A=\left(6.7^2+6.7^3+6.7^4+...+6.7^{101}\right)-\left(6.7+6.7^2+6.7^3+...+6.7^{100}\right)\)
\(6A=6.7^{101}-6.7=6\left(7^{101}-7\right)\Leftrightarrow A=7^{101}-7\)
vậy \(A=7^{101}-7\)
ta có : \(B=6.5-6.5^2+6.5^3-...+6.5^{99}-6.5^{100}\)
\(\Rightarrow5B=5\left(6.5-6.5^2+6.5^3-...+6.5^{99}-6.5^{100}\right)\)
\(5B=6.5^2-6.5^3+6.5^4-...+6.5^{100}-6.5^{101}\)
\(\Rightarrow5B+B=6B=6.5^2-6.5^3+6.5^4-...+6.5^{100}-6.5^{101}+6.5-6.5^2+6.5^3-...+6.5^{99}-6.5^{100}\)
\(6B=6.5-6.5^{101}=6.\left(5-5^{101}\right)\Leftrightarrow B=5-5^{101}\)
vậy \(B=5-5^{101}\)
MV
14 tháng 8 2017
\(A=6\cdot7+6\cdot7^2+6\cdot7^3+...+6\cdot7^{100}\\ =6\cdot\left(7+7^2+7^3+...+7^{100}\right)\\ =\left(7-1\right)\cdot\left(7+7^2+7^3+...+7^{100}\right)\\ =\left(7-1\right)\cdot7+\left(7-1\right)\cdot7^2+\left(7-1\right)\cdot7^3+...+\left(7-1\right)\cdot7^{100}\\ =7^2-7+7^3-7^2+7^4-7^3+...+7^{101}-7^{100}\\ =7^{101}-7=7\cdot\left(7^{100}-1\right)\)
\(B=6\cdot5-6\cdot5^2+6\cdot5^3-...+6\cdot5^{99}-6\cdot5^{100}\\ =6\cdot\left(5-5^2+5^3-...+5^{99}-5^{100}\right)\\ =\left(5+1\right)\cdot\left(5-5^2+5^3-...+5^{99}-5^{100}\right)\\=\left(5+1\right)\cdot5-\left(5+1\right)\cdot5^2+\left(5+1\right)\cdot5^3-...+\left(5+1\right)\cdot5^{99}-5^{100}\\ =5^2+5-5^3-5^2+5^4+5^3+...+5^{100}+5^{99}-5^{101}-5^{100}\\ =5-5^{101}\\ =5\cdot\left(1-5^{100}\right)\)
\(=\dfrac{2^5\cdot5^3}{2^6\cdot5^2}=\dfrac{5}{2}\)