Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{25.28}\right)=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}.\frac{3}{14}=\frac{5}{14}\)
\(D=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\)
\(D=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+\frac{2}{90}\)
\(D=\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+\frac{2}{9.10}\)
\(D=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(D=2\left(\frac{1}{4}-\frac{1}{10}\right)=2\cdot\frac{3}{20}=\frac{3}{10}\)
\(E=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(E=\frac{5}{28}+\frac{1}{14}+\frac{1}{26}+...+\frac{1}{140}\)
\(E=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(E=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{25.28}\)
\(E=\frac{5}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(E=\frac{5}{3}\cdot\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}\cdot\frac{3}{14}=\frac{5}{14}\)
Tổng của 9 số tự nhiên đầu là:
1+2+3+4+5+6+7+8+9=45
Vậy phân số đó là: \(\frac{45}{10}\)= \(\frac{9}{2}\)
tk nhé
Tổng của 9 số tự nhiên đầu tiên là:
1+2+3+4+5+6+7+8+9=45
Vậy ta có phân số: \(\frac{45}{10}\)= \(\frac{9}{2}\)
tk nhé
\(\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+....+\frac{8}{10}+\frac{9}{10}\)
\(=\frac{1+2+3+4+5+6+7+8+9}{10}\)
\(=\frac{\left(1+9\right)+\left(2+8\right)+\left(+7\right)+\left(4+6\right)+5}{10}\)
\(=\frac{10+10+10+10+5}{10}\)
\(=\frac{45}{10}=\frac{9}{2}\)
k mk nha
\(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}\)
\(=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\)
\(=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\right)\)
\(=1+\frac{28}{5}\)
\(=\frac{33}{5}\)
Ta có:
a) \(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{9}{15}=1+1+\frac{9}{15}=1\frac{9}{15}\)
b)\(\frac{10}{18}+\frac{4}{9}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}\right)\)
\(=1+\frac{31}{128}=1\frac{31}{128}\)
a,(11/15+4/15)+(5/7+2/7)
=1+1
=2
b,5/9x(1/2+6/4)
=5/9x2
=10/9
c,1/2:(7/8+9/8)
=1/2:2
=1
d,(17/10-7/10)+1/2
=1+1/2
=3/2
\(\frac{7}{10}-\frac{2}{5}+\frac{3}{10}+1\frac{2}{5}=\left(\frac{7}{10}+\frac{3}{10}\right)+\left(1\frac{2}{5}-\frac{2}{5}\right)=1+1=2\)
\(\frac{10}{56}+\frac{10}{140}+...+\frac{10}{1400}\)
\(=\frac{5}{28}+\frac{5}{70}+...+\frac{5}{700}\)
\(=\frac{5}{3}\left(\frac{3}{28}+\frac{3}{70}+...+\frac{3}{700}\right)\)
\(=\frac{5}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{25.28}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(=\frac{5}{3}.\frac{3}{14}=\frac{5}{14}\)