Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}+\frac{1}{6}.\frac{1}{7}+\frac{1}{7}.\frac{1}{8}+\frac{1}{8}.\frac{1}{9}\)
\(=\frac{1.1}{2.3}+\frac{1.1}{3.4}+\frac{1.1}{4.5}+\frac{1.1}{5.6}+\frac{1.1}{6.7}+\frac{1.1}{7.8}+\frac{1.1}{8.9}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}\)
\(=\frac{7}{18}\)
= 1/2.3+1/3.4+1/4.5+...+1/8.9
= 1/2-1/3+1/3-1/4+...+1/8-1/9
= 1/2-1/9
= 7/18
\(A=\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+..........+\frac{1}{8}.\frac{1}{9}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{8.9}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.......+\frac{1}{8}-\frac{1}{9}=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\)
\(B=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+....+\frac{1}{110}=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+.....+\frac{1}{10.11}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-.....+\frac{1}{10}-\frac{1}{11}=\frac{1}{4}-\frac{1}{11}=\frac{7}{44}\)
\(\text{c,d cơ bản tự làm nha }\)
A=>1.1/2.3+1.1/3.4+1.1/4.5+1.1/5.6+1.11/6.7+.1/7.8+1.1/8.9
=>1/2.3+1/3.4+1/4.5+1/6.7+1/7.8+1/8.9
=>1/2-1/3-1/4-1/5-1/6-1/7-1/8-1/9
=>1/2-1/9=>9/18-2/18=>7/18
Vậy A= 7/18
Bài 1:
a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)
b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)
c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)
d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)
e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)
hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
Tìm từng bài một ở câu hỏi tương tự nha bạn .
trả lời nhé!