hình như đề sai

mình xin lỗi các bạn giải cho mk nhé

a) \(\frac{2004}{2005}=1-\frac{1}{2005}\);\(\frac{2005}{2006}=1-\frac{1}{2006}\)

Vì \(\frac{1}{2005}>\frac{1}{2006}\)=>\(1-\frac{1}{2005}< 1-\frac{1}{2006}\)=>\(\frac{2004}{2005}< \frac{2005}{2006}\)

(1/2003+1/2004+1/2005) / (2/2003+2/2004+2/2005)

= (1/2003+1/2004+1/2005) / 2(1/2003+1/2004+1/2005)

= 1/2

Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)\)

\(A=1-\frac{1}{128}=\frac{127}{128}\)

Ủng hộ mk nha ^_-

\(2\cdot A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\)

\(A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)\)

\(A=1-\frac{1}{128}=\frac{128}{128}-\frac{1}{128}=\frac{127}{128}\)

\(\frac{2005}{2006}>\frac{2006}{2007}\)

\(\frac{2005}{2006}=0,9995014955...........\)

\(\frac{2006}{2007}=0,99955017439..............\)

mình ko ghi lại đề nha

a) \(1-\frac{2004}{2005}=\frac{1}{2005}\);\(1-\frac{2005}{2006}=\frac{1}{2006}\)

ta thấy \(\frac{1}{2005}>\frac{1}{2006}\)

=> \(\frac{2004}{2005}< \frac{2005}{2006}\)

b)\(\frac{2007}{2006}-1=\frac{1}{2006};\frac{2006}{2005}-1=\frac{1}{2005}\)

ta thấy \(\frac{1}{2006}< \frac{1}{2005}\)

=>\(\frac{2007}{2006}< \frac{2006}{2005}\)

c) ta có \(\frac{1975}{2005}>\frac{1975}{2006}>\frac{1974}{2006}\)

=>\(\frac{1975}{2005}>\frac{1974}{2006}\)

a, 135 / 120 < 13/8

b, 3535/4848 > 5/8

c, 650650/480480 < 222222/144144

Tk cho mk nha!!!!

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