1) so sanh \(3^{2011}\)va \(9^{1006}\)
2)\(3^x.\left(\frac{2}{3}\right)^x=16^2\)
Những câu hỏi liên quan
3. Tìm x biết :
\(\frac{x-1}{2011}+\frac{x-2}{2010}+\frac{x-3}{2009}=\frac{x-4}{2008}\)
2. Tìm x nguyên biết :
\(1-3+3^2-3^3+...+\left(-3\right)^x=\frac{9^{1006}-1}{4}\)
\(3.\)
\(\frac{x-1}{2011}+\frac{x-2}{2010}+\frac{x-3}{2009}=\frac{x-4}{2008}\)
\(\Rightarrow\)\(\frac{x-1}{2011}-1+\frac{x-2}{2010}-1+\frac{x-3}{2009}-1-\frac{x-4}{2008}+1+2=0\)
\(\Rightarrow\)\(\frac{x-1}{2011}-\frac{2011}{2011}+\frac{x-2}{2010}-\frac{2010}{2010}+\frac{x-3}{2009}-\frac{2009}{2009}-\frac{x-4}{2008}+\frac{2008}{2008}=0\)
\(\Rightarrow\)\(\frac{x-2012}{2011}+\frac{x-2012}{2010}+\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Rightarrow\)\(x-2012\left(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\right)=0\)
\(\Rightarrow\)\(x=2012\)
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tìm so nguyen x biet: a) \(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+..........+\frac{1}{\left(2x-1\right)x\left(2x+1\right)}=\frac{49}{99}\)
b) 1-3+32-33+.........+(-3)x=\(\frac{9^{1006}-1}{4}\)
\(1-3+3^2-3^3+..+\left(-3\right)^x=\frac{9^{1006}-1}{4}\)
\(1-3+3^2+3^3+...+\left(-3\right)^x=\frac{9^{1006-1}}{4}\)
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
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\(1-3+3^2+3^3+.....+\left(-3\right)^x=\frac{9^{1006}-1}{4}\)
Tìm x
Tìm x nguyên biết
\(1-3+3^2-3^3+.....+\left(-3\right)^x=\frac{9^{1006}-1}{4}\)
D = \(\frac{1006-\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2011}\right)}{\frac{2}{3}+\frac{4}{5}+...+\frac{2010}{2011}}\) =?
\(D=\frac{1006-\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2011}\right)}{\frac{2}{3}+\frac{4}{5}+...+\frac{2010}{2011}}\)
\(D=\frac{1006-\left(1+\frac{1}{3}+...+\frac{1}{2011}\right)}{1-\frac{1}{3}+1-\frac{1}{5}+...+1-\frac{1}{2011}}\)
\(D=\frac{1006-\left(1+\frac{1}{3}+...+\frac{1}{2011}\right)}{1006-\left(1+\frac{1}{3}+...+\frac{1}{2011}\right)}=1\)
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1. Tinh a left(6^9.2^{10}+12^{10}right)+left(2^{19}.27^3+15.4^9.9^4right)2. So sanh A va B.a) Afrac{-2012}{4025};Bfrac{-1999}{3997}b) A3^{21};B2^{31}c) Afrac{2011}{1.2}+frac{2011}{3.4}+frac{2011}{5.6}+....+frac{2011}{1999.2000};Bfrac{2012}{1001}+frac{2012}{1002}+frac{2012}{1003}+....+frac{2012}{2000}
Đọc tiếp
1. Tinh a \(\left(6^9.2^{10}+12^{10}\right)+\left(2^{19}.27^3+15.4^9.9^4\right)\)
2. So sanh A va B.
a) \(A=\frac{-2012}{4025};B=\frac{-1999}{3997}\)
b) \(A=3^{21};B=2^{31}\)
c) \(A=\frac{2011}{1.2}+\frac{2011}{3.4}+\frac{2011}{5.6}+....+\frac{2011}{1999.2000};\)\(B=\frac{2012}{1001}+\frac{2012}{1002}+\frac{2012}{1003}+....+\frac{2012}{2000}\)
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
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sao bạn lại nhắn vớ va vớ vậy PHẠM ĐỨC PHÚC
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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
= 2
18
.39
(2+3.22+5)=19.218
.39
\(1-333+3^2-33^3+......+\left(-3\right)^x=\frac{9^{1006}-1}{4}\)
tim x