\(a)\) \(A=\frac{1}{199}-\frac{1}{199.198}-\frac{1}{198.197}-\frac{1}{197.196}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(A=\frac{1}{199}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{197.198}+\frac{1}{198.199}\right)\)

\(A=\frac{1}{199}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{197}-\frac{1}{198}+\frac{1}{198}-\frac{1}{199}\right)\)

\(A=\frac{1}{199}-\left(1-\frac{1}{199}\right)\)

\(A=\frac{1}{199}-1+\frac{1}{199}\)

\(A=\frac{-197}{199}\)

Chúc bạn học tốt ~ 

làm hộ mk lun câu b ik

a) \(A=\frac{1}{199}-\frac{1}{199.198}-\frac{1}{198.197}-\frac{1}{197.196}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(A=\frac{1}{199}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...\frac{1}{197.198}+\frac{1}{198.199}\right)\)

\(A=\frac{1}{199}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{197}-\frac{1}{198}+\frac{1}{198}-\frac{1}{199}\right)\)

\(A=\frac{1}{199}-(1-\frac{1}{199})\)

\(A=\frac{1}{199}-1+\frac{1}{199}\)

\(A=\frac{-197}{199}\)

Vậy \(A=\frac{-197}{199}\)

b) cũng thế 

Bài 1:

a) \(B=1-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}-...-\frac{2}{61.63}-\frac{2}{63.65}\)

\(B=1-\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{61.63}+\frac{2}{63.65}\right)\)

\(B=1-\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{61}-\frac{1}{63}+\frac{1}{63}-\frac{1}{65}\right)\)

\(B=1-\left(\frac{1}{3}-\frac{1}{65}\right)\)

\(B=1-\frac{62}{195}\)

\(B=\frac{133}{195}\)

b) \(C=1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-...-\frac{1}{95.100}\)

\(C=1-\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{95.100}\right)\)

\(C=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{95}-\frac{1}{100}\right)\)

\(C=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{100}\right)\)

\(C=1-\frac{1}{5}.\frac{19}{100}\)

\(C=1-\frac{19}{500}\)

\(C=\frac{481}{500}\)

bài 2 thì bn lm như bn Phùng Minh Quân nha!

Câu 1 : mình ko hiểu đề bài cho lắm ~.~ 

Câu 2 : 

Ta có : 

\(\left|\frac{1}{2}-x\right|\ge0\)

\(\Rightarrow\)\(A=10+\left|\frac{1}{2}-x\right|\ge10\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\left|\frac{1}{2}-x\right|=0\)

\(\Leftrightarrow\)\(\frac{1}{2}-x=0\)

\(\Leftrightarrow\)\(x=\frac{1}{2}\)

Vậy GTNN của \(A\) là \(10\) khi \(x=\frac{1}{2}\)

Chúc bạn học tốt ~ 

\(=1-\frac{62}{195}\)

\(=\frac{133}{195}\)

Ta có : \(A=\dfrac{1}{199}-\dfrac{1}{199.198}-\dfrac{1}{198.197}-\dfrac{1}{197.196}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(=\dfrac{1}{199}-\left(\dfrac{1}{199.198}+\dfrac{1}{198.197}+\dfrac{1}{197.196}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)

\(=\dfrac{1}{199}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{196.197}+\dfrac{1}{197.198}+\dfrac{1}{198.199}\right)\)

\(=\dfrac{1}{199}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...-\dfrac{1}{198}+\dfrac{1}{198}-\dfrac{1}{199}\right)\)

\(=\dfrac{1}{199}-\left(1-\dfrac{1}{199}\right)\)

\(=\dfrac{1}{199}-\dfrac{198}{199}=\dfrac{-197}{199}\)

~ Học tốt ~

1-1/3-1/65

\(A=1-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}-...-\frac{2}{63.65}\)

\(A=1-\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{63-65}\right)\)

\(A=1-\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{63}-\frac{1}{65}\right)\)

\(A=1-\left(\frac{1}{3}-\frac{1}{65}\right)\)

\(A=1-\frac{62}{195}\)

\(A=\frac{133}{195}\)

Ta có: \(B=1-\dfrac{2}{3.5}-\dfrac{2}{5.7}-\dfrac{2}{7.9}-...-\dfrac{2}{61.63}-\dfrac{2}{63.65}\)

\(=1-\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{61.63}+\dfrac{2}{63.65}\right)\)

\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{63}-\dfrac{1}{65}\right)\)

\(=1-\left(\dfrac{1}{3}-\dfrac{1}{65}\right)\)

\(=1-\dfrac{62}{195}=\dfrac{133}{195}.\)

Vậy \(B=\dfrac{133}{195}.\)

6

=7

  A= \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+\(\dfrac{1}{7.9}\)+...+\(\dfrac{1}{97.99}\)

2A= 1 - \(\dfrac{1}{3}\)+\(\dfrac{1}{3}\) - \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\) - \(\dfrac{1}{9}\)+...+\(\dfrac{1}{97}\)-\(\dfrac{1}{99}\)

2A= 1-\(\dfrac{1}{99}\)

2A= \(\dfrac{98}{99}\)

  A= \(\dfrac{98}{99}\) : 2

A=\(\dfrac{49}{99}\)