E
1
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
25 tháng 9 2018
a, A= 5x415 x 99 - 4x320x 89 = 5x230x318-22x 227x320
=229x318(2x5-32)=229x318
B=5x29x619-7x229x276=5x29x219x3197x229x318
=238x318(5x3-7x2)=228-318
->A:B= 229x318:228:318=2
DP
8 tháng 7 2017
\(b,\)Đặt \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37\cdot38\cdot39}\)
\(B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{37.38\cdot38}\)
\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(2B=\frac{1}{1.2}-\frac{1}{38.39}\)
\(\Rightarrow B=\frac{\left(\frac{1}{1.2}-\frac{1}{38.39}\right)}{2}=\frac{185}{741}\)
MC
14 tháng 7 2018
\(1)C=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{162}\)
\(3C=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{54}\)
\(3C-C=\left(1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{54}\right)-\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{162}\right)\)
\(2C=1-\dfrac{1}{162}\)
\(2C=\dfrac{161}{162}\)
\(C=\dfrac{161}{162}.\dfrac{1}{2}\)
\(C=\dfrac{161}{324}\)
\(2)A=\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+\dfrac{1}{128}+\dfrac{1}{512}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+\dfrac{1}{128}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+\dfrac{1}{128}\right)-\left(\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+\dfrac{1}{128}+\dfrac{1}{512}\right)\)
\(A=1-\dfrac{1}{512}=\dfrac{511}{512}\)
NH
0
21 tháng 9 2019
a.
\(A=5.\left(2^2\right)^{15}.\left(3^2\right)^9-2^2.3^{20}.\left(2^3\right)^9=5.2^{30}.3^{18}-2^2.3^{20}.2^{27}\)
\(=5.2^{30}.3^{18}-3^{20}.2^{29}=2^{29}.3^{18}.\left(5.2-3^2\right)=2^{29}.3^{18}\)
\(B=5.2^9.\left(2.3\right)^{19}-7.2^{29}.\left(3^3\right)^6=5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}=5.2^{28}.3^{19}-7.2^{29}.3^{18}\)
\(=2^{28}.3^{18}.\left(5.3-7.2\right)=2^{28}.3^{18}\)
=> \(A:B=\left(2^{29}.3^{18}\right):\left(2^{28}.3^{18}\right)=\frac{\left(2^{29}.3^{18}\right)}{\left(2^{28}.3^{18}\right)}=2\)
b. kiểm tra lại đề bài nhé
DV
0
A=\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{243}+\frac{1}{729}\)
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^5}+\frac{1}{3^6}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^4}+\frac{1}{3^5}\)
Lấy 3A - A ta được : (\(1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^4}+\frac{1}{3^5}\) ) - (\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^5}+\frac{1}{3^6}\))
2A = 1 - \(\frac{1}{3^6}\)
=> A = \(\frac{1-\frac{1}{3^6}}{2}=\frac{364}{729}\)