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
MV
29 tháng 10 2017
\(A=\dfrac{12^{15}\cdot3^4-4^5\cdot3^9}{27^3\cdot2^{10}-32^3\cdot3^9}\\ =\dfrac{\left(2^2\cdot3\right)^{15}\cdot3^4-\left(2^2\right)^5\cdot3^9}{\left(3^3\right)^3\cdot2^{10}-\left(2^5\right)^3\cdot3^9}\\ =\dfrac{2^{30}\cdot3^{15}\cdot3^4-2^{10}\cdot3^9}{3^9\cdot2^{10}-2^{15}\cdot3^9}\\ =\dfrac{3^9\cdot2^{10}\left(2^{20}\cdot3^{10}\right)}{3^9\cdot2^{10}\left(1-2^5\right)}\\ =\dfrac{\left(2^2\right)^{10}\cdot3^{10}}{1-32}\\ =\dfrac{\left(2^2\cdot3\right)^{10}}{-31}\\ =\dfrac{-12^{10}}{31}\)
\(B=\dfrac{3}{1^2\cdot2^2}+\dfrac{5}{2^2\cdot3^2}+...+\dfrac{99}{49^2\cdot50^2}\\ =\dfrac{2^2-1^2}{1^2\cdot2^2}+\dfrac{3^2-2^2}{2^2\cdot3^2}+...+\dfrac{50^2-49^2}{49^2\cdot50^2}\\ =\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{49^2}-\dfrac{1}{50^2}\\ =1-\dfrac{1}{2500}\\ =\dfrac{2499}{2500}\)
16 tháng 4 2018
Ta có :
\(A=\frac{3^9-2^3.3^7+2^{10}.3^2-2^{13}}{3^{10}-2^2.3^7+2^{10}.3^3-2^{12}}\)
\(A=\frac{3^7\left(3^2-2^3\right)+2^{10}\left(3^2-2^3\right)}{3^7\left(3^3-2^2\right)+2^{10}\left(3^3-2^2\right)}\)
\(A=\frac{\left(3^2-2^3\right)\left(3^7+2^{10}\right)}{\left(3^3-2^2\right)\left(3^7+2^{10}\right)}\)
\(A=\frac{3^2-2^3}{3^3-2^2}\)
\(A=\frac{9-8}{27-4}\)
\(A=\frac{1}{23}\)
Vậy \(A=\frac{1}{23}\)
Chúc bạn học tốt ~
20 tháng 4 2018
\(B=\frac{3^9-2^3\cdot3^7+2^{10}\cdot3^2-2^{13}}{3^{10}-2^2\cdot3^7+2^{10}\cdot3^3-2^{12}}\)
\(B=\frac{1-2\cdot1+1\cdot1-2}{3-1\cdot1+1\cdot3-1}\)
\(B=\frac{1-2+1-2}{3-1+3-1}\)
\(B=\frac{-1+\left(-1\right)}{2+2}\)
\(B=\frac{-2}{4}\)
\(\Rightarrow B=\frac{-1}{2}\)
NL
0
D
19 tháng 5 2017
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}+\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{12}}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{12}}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6-2^{12}.3^5}-\frac{2^{12}.3^{10}-2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{2^{12}.\left(3^5-3^4\right)}{2^{12}.\left(3^6-3^5\right)}-\frac{2^{12}.3^{10}-2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{3^5-3^4}{3^6-3^5}-\frac{2^{12}.3^{10}.\left(1-5\right)}{2^{13}.3^{12}}\)
\(=\frac{162}{486}-\frac{2^{12}.3^{10}.\left(-4\right)}{2^{13}.3^{10}.3^2}=\frac{1}{3}-\frac{2^{14}.3^{10}.\left(-1\right)}{2^{13}.3^{10}.9}\)
\(=\frac{1}{3}-\frac{2.1.\left(-1\right)}{1.1.9}=\frac{1}{3}-\frac{2}{9}=\frac{1}{9}\)
20 tháng 5 2017
eo ôi t làm rồi mà bị xoá :v thôi t hướng dẫn :v
Tạc TS và MS ra rồi gộp và triệt tiêu :) nếu k lm đc ibx t làm cho :)
N
20 tháng 7 2019
mk doan la` de sai, sua: \(\frac{3^9-2^3.3^7+2^{10}.3^2-2^{13}}{3^{10}-2^2.3^7+2^{10}.3^3-2^{12}}\)
\(=\frac{3^7.\left(3^2-2^3\right)+2^{10}.\left(3^2-2^3\right)}{3^7.\left(3^3-2^2\right)+2^{10}.\left(3^3-2^2\right)}=\frac{3^7+2^{10}}{\left(3^7+2^{10}\right).24}=\frac{1}{24}\)
21 tháng 8 2020
\(5^6:5^3+3^3.3^2\)
\(=5^2+3^3.3^2\)
\(=5^2+3^5\)
\(=25+243\)
\(=268\)
LT
1
28 tháng 10 2016
\(a=\frac{16^3.3^{10}+120.6^9}{4^6+3^{12}+6^{11}}=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6+3^{12}+\left(2.3\right)^{11}}=\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}+3^{12}+2^{11}.3^{11}}=\)
9^3.3^2/3^9.2022 = 1002618.50877
TL
1 474 038
HT