???

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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: $32<2024$ 

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

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

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

$Vậy:A<\frac{2023}{2024}$ 

\(x+\left(x+1\right)+\left(x+2\right)+....+\left(x+2003\right)=2004\)

\(\Leftrightarrow x+\left(x+x+....+x\right)+\left(1+2+....+2003\right)=2004\)

\(\Leftrightarrow x+2003x+2007006=2004\)

\(\Leftrightarrow2004x=2004-2007006\)

\(\Leftrightarrow2004x=-2005002\)

\(\Leftrightarrow x=-\frac{2001}{2}=-1000,5\)

MÌNH QUÊN cách lm rùi

đặt B=99/1+99/2+...+1/99

=1+(98/2+1)+(97/3+1)+...+(1/99+1)

=100/100+100/2+...+100/99

=100.(1/2+1/3+...+1/100)

=>A=(1/2+1/3+...+1/100):[100.(1/2+1/3+...+1/100)]

A=1:100=1/100

hok tốt nha

\(\dfrac{2003x1999-2003x999}{2004x999+1004}\)

\(=\dfrac{2003x\left(1999-999\right)}{2004x\left(1000-1\right)+1004}\)

\(=\dfrac{2003x1000}{2004x1000-2004+1004}\)

\(=\dfrac{2003x1000}{2004x1000-1000}\)

\(=\dfrac{2003x1000}{\left(2004-1\right)x1000}\)

\(=\dfrac{2003x1000}{2003x1000}=1\)

= 1 nhé

cảm ơn bạn nha