???

Are you sure?

nói chung là :

làm đc vế a là làm đc vế b

$A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}$ 

$=\frac{16}{32}+\frac{8}{32}+\frac{4}{32}+\frac{2}{32}+\frac{1}{32}$ 

$=\frac{31}{32}$ 

Xét: $1-A=1-\frac{31}{32}=\frac{1}{32}$ 

Xét: $1-\frac{2023}{2024}=\frac{1}{2024}$ 

Vì: $32<2024$ 

$\Rightarrow \frac{1}{32}>\frac{1}{2024}$ 

$\Rightarrow 1-\frac{1}{32}<1-\frac{1}{2024}$ 

$\Rightarrow \frac{31}{32}<\frac{2023}{2024}$ 

$Vậy:A<\frac{2023}{2024}$ 

127^2 + 146 x 126 + 73^2 

= 127^2 + 2 x 73 x 126 + 73 x 73

= 127^2 + 73 x ( 2 x126 + 73 )

=......

rồi sau đo tinh binh thuong mk chi co the giup vay thoi

2004/2003

day nha ban

2004/2003

k nha

link nee

https://olm.vn/hoi-dap/detail/47059131770.html

nho ti

ck cho mk

a)2012/2013<2013/2014

b)2003x........< 2007x......

cho mình nha

\(\frac{2004}{2005}\cdot\frac{2005}{2003}\cdot\frac{2003}{2004}=\frac{2004\cdot2005\cdot2003}{2005\cdot2003\cdot2004}=1\)

\(\frac{2004x2005x2003}{2005x2003x2004}\)

Sau đó thì bạn chỉ cần rút gọn thui

Đáp án là 1 nhé

a: \(=5^{2003}\left(5^2-5+1\right)\)

\(=5^{2003}\cdot21⋮7\)

\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{3004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)

\(\Rightarrow P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)

\(\Rightarrow P=\frac{1}{5}-\frac{2}{3}\)

\(\Rightarrow P=\frac{3}{15}-\frac{10}{15}\)

\(\Rightarrow P=\frac{-7}{15}\)

Vậy \(P=\frac{-7}{15}\)

D = \(\frac{2^{2004}+1}{2^{2003}+1}\)=\(\frac{2^{2003}+2}{2^{2004}+2}\)

C = \(\frac{2^{2005}+3}{2^{2006}+3}\)= \(\frac{2^{2005}+2}{2^{2006}+2}\)

 Vậy C>D

mình chuyển 1 hạng tử của 3 từ bên d sang c nên ta được pt như trên