Bài 3:

\(a,\dfrac{x-1}{10}+\dfrac{x-1}{11}=\dfrac{x-1}{12}+\dfrac{x-1}{13}\)

\(\Rightarrow\dfrac{x-1}{10}+\dfrac{x-1}{11}-\dfrac{x-1}{12}-\dfrac{x-1}{13}=0\)

\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\right)=0\)

Mà \(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\ne0\)

\(\Rightarrow x-1=0\Rightarrow x=1\)

Vậy x = 1

b, \(\dfrac{x-2000}{10}+\dfrac{x-1999}{9}=\dfrac{x-1998}{8}+\dfrac{x-1997}{7}\)

\(\Rightarrow\dfrac{x-2000}{10}+1+\dfrac{x-1999}{9}+1=\dfrac{x-1998}{8}+\dfrac{x-1997}{7}+1\)

\(\Rightarrow\dfrac{x-1990}{10}+\dfrac{x-1990}{9}-\dfrac{x-1990}{8}-\dfrac{x-1990}{7}=0\)

\(\Rightarrow\left(x-1990\right)\left(\dfrac{1}{10}+\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{7}\right)=0\)

Mà \(\dfrac{1}{10}+\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{7}\ne0\)

\(\Rightarrow x-1990=0\Rightarrow x=1990\)

a) 

\(1-\frac{1998}{1999}=\frac{1}{1999}\)

\(1-\frac{1999}{2000}=\frac{1}{2000}\)

Vì \(\frac{1}{1999}>\frac{1}{2000}\)nên \(\frac{1998}{1999}< \frac{1999}{2000}\)

b) Ta có :

\(\frac{1999}{2001}< 1\)

\(\frac{12}{11}>1\)

Nên \(\frac{1999}{2001}< \frac{12}{11}\)

c) 

\(1-\frac{13}{27}=\frac{14}{27}\)

\(1-\frac{27}{41}=\frac{14}{41}\)

Vì \(\frac{14}{27}>\frac{14}{41}\)nên \(\frac{13}{27}< \frac{27}{41}\)

d) 

Ta có phân số trung gian là \(\frac{23}{45}\).

Ta có : \(\frac{23}{47}< \frac{23}{45}\) ; \(\frac{24}{45}>\frac{23}{45}\)

Nên \(\frac{23}{47}< \frac{24}{45}\)

có ai trả lời mik ko 

ai trả lời được mik liền

Ta có : \(1=\frac{1998}{1999}+\frac{1}{1999}\)

         \(1=\frac{1}{2000}+\frac{1999}{2000}\)

Mà \(\frac{1}{2000}< \frac{1}{1999}\)

Nên \(\frac{1999}{2000}>\frac{1998}{1999}\)

Ta có:\(\frac{1}{11}>\frac{1}{20};\frac{1}{12}>\frac{1}{20};\frac{1}{13}>\frac{1}{20};....;\frac{1}{19}>\frac{1}{20}\)

\(\Rightarrow\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)(Có 10 phân số \(\frac{1}{20}\))

\(\Rightarrow\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{10}{20}\)\(\Leftrightarrow S>\frac{10}{20}\)

Mà \(\frac{10}{20}=\frac{1}{2}\)nên

\(\Rightarrow S>\frac{1}{2}\)

\(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}+...+\frac{1}{20}\)(10 số \(\frac{1}{20}\))

=\(\frac{1}{20}.10=\frac{1}{2}\)

vậy S>1/2

vậy 1/2 = 1/2 vì ( S = 1/20 )

Ta có các phân số 1/11 ; 1/12 ; 1/13 ; 1/14 ; 1/15 ; 1/16 ; 1/17 ; 1/18 ; 1/19 đều lớn hơn 1/20

Do đó : 1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16 + 1/17 + 1/18 + 1/19 + 1/20 > 1/20 + 1/20 + ;...+ 1/20 ( có 10 phân số 1/20 )

1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1 /16 + 1/17 + 1/18 + 1/19 + 1/20 > 10/20

1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1 /16 + 1/17 + 1/18 + 1/19 + 1/20 > 1/2

Vậy : S > 1/2

S>\(\frac{1}{2}\)

s>1/2

\(S=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\)

\(>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)(10 số hạng) 

\(=10.\frac{1}{20}=\frac{1}{2}\).

Vậy \(S>\frac{1}{2}\).

ta thấy: 1/11;1/12;1/13;...;1/19;1/20 đều >1/20

=>1/11+1/12+...1/19+1/20>1/20+1/20...+1/20

1/11+1/12+...1/19+1/20>10/20

1/11+1/12+...1/19+1/20>1/2 vậy S>1/2