NH
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
DV
0
BM
0
31 tháng 7 2020
1. Ta có :
\(4A=\frac{2^2\left(2^{18}-3\right)}{2^{20}-3}=\frac{2^{20}-12}{2^{20}-3}=\frac{2^{20}-3-9}{2^{20}-3}=\frac{2^{20}-3}{2^{20}-3}-\frac{9}{2^{20}-3}=1-\frac{9}{2^{20}-3}\)
\(4B=\frac{2^2\left(2^{20}-3\right)}{2^{22}-3}=\frac{2^{22}-12}{2^{22}-3}=\frac{2^{22}-3-9}{2^{22}-3}=\frac{2^{22}-3}{2^{22}-3}-\frac{9}{2^{22}-3}=1-\frac{9}{2^{22}-3}\)
Vì \(2^{20}-3< 2^{22}-3\)
\(\Leftrightarrow\frac{9}{2^{20}-3}>\frac{9}{2^{22}-3}\)
\(\Leftrightarrow1-\frac{9}{2^{20}-3}< 1-\frac{9}{2^{22}-3}\)
\(\Leftrightarrow4A< 4B\)
\(\Leftrightarrow A< B\)
Vậy...
b/ Tương tự
HM
0
EC
0
LG
0
VM
6
6 tháng 4 2015
\(\frac{3}{2}.A=\frac{3}{4}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+...+\left(\frac{3}{2}\right)^{2013}\)
\(\Rightarrow\frac{3}{2}.A-A=\frac{3}{4}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+...+\left(\frac{3}{2}\right)^{2013}-\left(\frac{1}{2}+\frac{3}{2}+\left(\frac{3}{2}\right)^2+...+\left(\frac{3}{2}\right)^{2012}\right)\)
\(\Rightarrow\frac{1}{2}.A=\frac{3}{4}+\left(\frac{3}{2}\right)^{2013}-\frac{1}{2}-\frac{3}{2}=\left(\frac{3}{2}\right)^{2013}-\frac{5}{4}\Rightarrow A=2.\left(\frac{3}{2}\right)^{2013}-\frac{5}{2}\)
\(B-A=\frac{1}{2}.\left(\frac{3}{2}\right)^{2013}-2.\left(\frac{3}{2}\right)^{2013}+\frac{5}{2}=-\left(\frac{3}{2}\right)^{2014}+\frac{5}{2}\)
NC
0
Lời giải:
$A=\frac{1}{2}+\frac{3}{2}+(\frac{3}{2})^2+(\frac{3}{2})^3+....+(\frac{3}{2})^{2012}$
$\frac{3}{2}A=\frac{3}{4}+(\frac{3}{2})^2+(\frac{3}{2})^3+(\frac{3}{2})^4+...+(\frac{3}{2})^{2013}$
$\Rightarrow \frac{3}{2}A-A=(\frac{3}{2})^{2013}+\frac{3}{4}-\frac{1}{2}-\frac{3}{2}$
$\Rightarrow \frac{1}{2}A=(\frac{3}{2})^{2013}-\frac{5}{4}$
$A=2(\frac{3}{2})^{2013}-\frac{5}{2}$
$\Rightarrow B-A=(\frac{3}{2})^{2013}-2(\frac{3}{2})^{2013}+\frac{5}{2}=\frac{5}{2}-(\frac{3}{2})^{2013}$