A=\(\frac{19}{20}\),

sao ra vay ban minh muoc cach giai bai

Ta có:  1/1.2 = 1- 1/2 

           1/3.4 = 1/3 - 1/4 

          ...............

             1/19.20 = 1/19 - 1/20 

Cộng vế với vế ta đc: 

    A = 1- 1/20 = 19/20   

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{17.18}+\frac{1}{19.20}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{17}-\frac{1}{18}+\frac{1}{19}-\frac{1}{20}\)

\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{17}+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+..+\frac{1}{18}+\frac{1}{20}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{18}+\frac{1}{20}\right)\)

\(A=\left(1+\frac{1}{2}+...+\frac{1}{20}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{9}+\frac{1}{10}\right)\)

\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}+\frac{1}{20}\)

\(\frac{A}{B}=1\)

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{19.20}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)

\(\Rightarrow A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{20}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{20}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)

\(\Rightarrow A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}=B\)

\(\Rightarrow\frac{A}{B}=1\)

a)Ta có:

 A= 1/1.2+1/2.3+1/3.4+.....+1/99.100

=1-1/2+1/2-1/3+...+1/99-1/100

=1-1/100

=99/100

b)Ta có:

B= 1/11+1/12+1/13+1/14+1/15+...+1/50

=(1/11+1/50)+(1/12+1/49)+...+(1/30+1/31)

=61/11.50+61/12.49+...+61/30.31

=61.(1/11.50+1/12.49+...+1/30.31)

Mình xin lỗi chỉ làm được đến đây vì dạng tính B mình không tốt lắm ◕◡◕

\(B=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{50}\right)>\left(\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\)=> \(B>\frac{20}{30}+\frac{20}{50}=\frac{2}{3}+\frac{2}{5}=\frac{16}{15}>1\)

mà \(A=\frac{99}{100}

ssssssssss

Tính tử số:

\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{19.20}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{20}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{20}\right)\)

\(=1+\frac{1}{2}+...+\frac{1}{20}-\left(1+\frac{1}{2}+...+\frac{1}{10}\right)\)

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

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

a)Đặt \(A=\dfrac{6}{1.4}+\dfrac{6}{4.7}+\dfrac{6}{7.10}+...+\dfrac{6}{97.100}\)

\(3a=3-\dfrac{3}{4}+\dfrac{3}{4}-\dfrac{3}{7}+\dfrac{3}{7}-\dfrac{3}{10}+...+\dfrac{3}{97}-\dfrac{3}{100}\)

\(=3-\dfrac{3}{100}\)

\(=\dfrac{297}{100}\)

b)Đặt \(B=\dfrac{4}{1.3}+\dfrac{16}{3.5}+\dfrac{36}{5.7}+...+\dfrac{9604}{97.99}\)

\(=2b=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\)

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

\(2b=2-\dfrac{2}{99}=\dfrac{198}{99}-\dfrac{2}{99}=\dfrac{196}{99}\)

c) Tương tự! Bạn tự làm nhé!