\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+...+\dfrac{1}{\left(x+1999\right)\left(x+2000\right)}=\dfrac{1}{x+200}=\dfrac{1}{5}\)\(\Rightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+...+\dfrac{1}{x+1999}-\dfrac{1}{x-2000}=\dfrac{1}{x+200}=\dfrac{1}{5}\)

\(\Rightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2000}=\dfrac{1}{x+200}=\dfrac{1}{5}\)

Đề này sai nhé,hình như thừa dữ kiện đề r

BT 1 Sai De ui

\(M=1+\frac{1}{199}+1+\frac{2}{198}+1+....+\frac{198}{2}+1=\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+....+\frac{200}{2}\)

  \(=200.\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)=200 T

\(S=\frac{T}{200T}=\frac{1}{200}\)

jups mình với đang cần gấp

lớp 2 chưa học phân số

@@ gửi ít thôi bạn

a) Vì \(721< 834\Rightarrow\frac{5}{721}>\frac{5}{834}\)

b) Ta có \(\frac{4}{37}< \frac{5}{37}< \frac{5}{36}\Rightarrow\frac{4}{37}< \frac{5}{36}\)

c) Ta có \(\frac{1994}{1995}=1-\frac{1}{1995}\)

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

Vì \(\frac{1}{1995}>\frac{1}{2000}\Rightarrow1-\frac{1}{1995}< 1-\frac{1}{2000}\Rightarrow\frac{1994}{1995}< \frac{1999}{2000}\)

d) Ta có :\(\frac{489}{487}=1+\frac{2}{487}\)

\(\frac{487}{485}=1+\frac{2}{485}\)

Vì \(\frac{2}{485}>\frac{2}{487}\Rightarrow1+\frac{2}{485}>1+\frac{2}{487}\Rightarrow\frac{489}{487}>\frac{487}{485}\)

e) Ta có : \(\frac{123.125+119}{124.125-177}=\frac{123.125+119}{\left(123+1\right).125-177}=\frac{123.125+119}{123.125+125-177}=\frac{123.125+119}{123.125-52}\)

\(=\frac{123.125-52+171}{123.125-52}=1+\frac{171}{123.125-52}>1\)

f) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{199}-\frac{1}{200}=1-\frac{1}{200}< 1\)