1) so sanh \(3^{2011}\)va \(9^{1006}\)
2)\(3^x.\left(\frac{2}{3}\right)^x=16^2\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(3.\)
\(\frac{x-1}{2011}+\frac{x-2}{2010}+\frac{x-3}{2009}=\frac{x-4}{2008}\)
\(\Rightarrow\)\(\frac{x-1}{2011}-1+\frac{x-2}{2010}-1+\frac{x-3}{2009}-1-\frac{x-4}{2008}+1+2=0\)
\(\Rightarrow\)\(\frac{x-1}{2011}-\frac{2011}{2011}+\frac{x-2}{2010}-\frac{2010}{2010}+\frac{x-3}{2009}-\frac{2009}{2009}-\frac{x-4}{2008}+\frac{2008}{2008}=0\)
\(\Rightarrow\)\(\frac{x-2012}{2011}+\frac{x-2012}{2010}+\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Rightarrow\)\(x-2012\left(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\right)=0\)
\(\Rightarrow\)\(x=2012\)
Vì \(\frac{9^{1006}-1}{4}\) là số chẵn nên x là số lẻ
\(\Rightarrow\left(-3\right)^x=-3^x\)
Đặt A=1-3+32-33+...-3x
3A=3-32+33-34+...+3x+1
3A+A=[3-32+33-34+...+3x+1] -[1-3+32-33+...-3x]
4A=3x+1-1
\(A=\frac{3^{x+1}-1}{4}=\frac{9^{1006}-1}{4}=\frac{\left(3^2\right)^{1006}-1}{4}=\frac{3^{1012}-1}{4}\)
=>x+1=2012
=>x=2012-1=2011
vậy x=2011
\(D=\frac{1006-\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2011}\right)}{\frac{2}{3}+\frac{4}{5}+...+\frac{2010}{2011}}\)
\(D=\frac{1006-\left(1+\frac{1}{3}+...+\frac{1}{2011}\right)}{1-\frac{1}{3}+1-\frac{1}{5}+...+1-\frac{1}{2011}}\)
\(D=\frac{1006-\left(1+\frac{1}{3}+...+\frac{1}{2011}\right)}{1006-\left(1+\frac{1}{3}+...+\frac{1}{2011}\right)}=1\)
1/ (69.210+1210)+(219.273+15.49.94) = 29.39.210+310.220+219.39+5.3.218.38 = 219.39+310.220+219.39+5.218.39
= 218.39(2+3.22+5)=19.218.39
1, Ta có : \(9^{1006}=\left(3^2\right)^{1006}=3^{2012}\)
Vì \(2011< 2012\)
\(\Rightarrow3^{2011}< 3^{2012}\)
bài 2 bạn tự làm nha