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

\(B=\dfrac{1}{1}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{5}+\dfrac{1}{5}\cdot\dfrac{1}{7}+...+\dfrac{1}{97}\cdot\dfrac{1}{99}\)

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

\(B=\dfrac{1}{1}-\dfrac{1}{99}\)

\(B=\dfrac{99}{99}-\dfrac{1}{99}\)

\(B=\dfrac{98}{99}\)

#YVA

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

B=\(\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right):2\)

B=\(\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{97}-\dfrac{1}{99}\right):2\)

B=\(\left(\dfrac{1}{1}-\dfrac{1}{99}\right):2\)

B=\(\dfrac{98}{99}:2\)

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

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

⇔ \(2A=2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right)\)

⇔ 2A = \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\)

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

⇔ 2A = \(\dfrac{1}{3}-\dfrac{1}{99}\)

⇔ 2A = \(\dfrac{32}{99}\)

⇔ A = \(\dfrac{32}{99}:2\)

⇔ A = \(\dfrac{16}{99}\)

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

\(\Leftrightarrow2A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+......+\dfrac{2}{97.99}\)

\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+.....+\dfrac{1}{97}-\dfrac{1}{99}\)

\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{99}\)

\(\Leftrightarrow2A=\dfrac{32}{99}\)

\(\Leftrightarrow A=\dfrac{16}{99}\)

a: \(A=\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\cdots+\frac{2}{99\cdot100}\)

\(=2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{99\cdot100}\right)\)

\(=2\left(\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{99}-\frac{1}{100}\right)\)

\(=2\left(\frac12-\frac{1}{100}\right)=1-\frac{2}{100}=\frac{98}{100}=\frac{49}{50}\)

b: \(A=\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^{20}}\)

=>2A=\(1+\frac12+\cdots+\frac{1}{2^{19}}\)

=>2A-A=\(1+\frac12+\cdots+\frac{1}{2^{19}}-\frac12-\frac{1}{2^2}-\cdots-\frac{1}{2^{20}}\)

=>A=\(1-\frac{1}{2^{20}}\)

=>A<1

c: \(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\cdots+\frac{1}{97\cdot99}\)

\(=\frac12\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\cdots+\frac{2}{97\cdot99}\right)\)

\(=\frac12\left(\frac13-\frac15+\frac15-\frac17+\cdots+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac12\left(\frac13-\frac{1}{99}\right)=\frac12\cdot\frac{32}{99}=\frac{16}{99}\)

\(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{95\cdot97}+\dfrac{2}{97\cdot99}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{99}\)

\(=\dfrac{1}{3}-\dfrac{1}{99}\)

\(=\dfrac{33}{99}-\dfrac{1}{99}\)

\(=\dfrac{32}{99}\)

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{95.97}+\dfrac{2}{97.99}\)

\(\Rightarrow\dfrac{2}{3}-\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{2}{7}+...+\dfrac{2}{95}-\dfrac{2}{97}+\dfrac{2}{97}-\dfrac{2}{99}\)

\(\Rightarrow\dfrac{2}{3}-\dfrac{2}{99}=\dfrac{64}{99}\)

chúc bạn học tốt

\(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)

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

\(M=2.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)

\(M=2.\dfrac{32}{99}\)

\(M=\dfrac{64}{99}\)

http://vietjack.com/giai-sach-bai-tap-toan-6/bai-95-trang-28-sach-bai-tap-toan-6-tap-2.jsp

\(B=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\\ B=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\\ B=\dfrac{1}{1}-\dfrac{1}{101}\\ B=\dfrac{101}{101}-\dfrac{1}{101}\\ B=\dfrac{100}{101}\)

a, đặt đề bài là A

Ta có : A=( 1-1/2+1/2-1/3+...+1/9-1/10).(x-1)+1/10.x=x-9/10

= (1-1/10).(x-1)+1/10.x

= 9/10 .( x-1 )+1/10.x

=1.x-9/10

nên x= 0 hoặc 1

với -1 nữa nha

chữ xấu (thông cảm) ;-;

\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{101}\right)=2\cdot\dfrac{98}{303}=\dfrac{196}{303}\)

= 2/3 . 2/5 + 2/5 . 2/7 + ... + 2/99 . 2/101

= 2/3 - 2/5 + 2/5 - 2/7 + ... + 2/99 - 2/101

= 2/3 - 2/101

= 196/303