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ta tính nhanh như sau
1/3+1/12=5/12
vậy tổng trên =5/12 nhớ k hoặc kết bạn vs minh nha
\(x+\frac{1}{2}+x+\frac{1}{4}+x+\frac{1}{8}+x+\frac{1}{16}=\frac{23}{16}=>4x=\frac{23}{16}-\frac{15}{16}=\frac{1}{2}=>x=\frac{1}{8}\)
(xx4)+1/2+1/4+1/8+1/16=23/16
xx4=23/16-(1/2+1/4+1/8+1/16)
xx4=1/2
x=1/2:4
x=1/8
Đúng 100% nhé Nhớ k cho mình đấy
1, \(\frac{7}{12}=\frac{1}{12}+\frac{2}{12}+\frac{4}{12}=\frac{1}{12}+\frac{1}{6}+\frac{1}{3}\)
2, \(S=\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{19x20}\)
\(=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{19}-\frac{1}{20}\right)\)
\(=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
Chúc bạn học tốt.
\(a.\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)
\(=\frac{1}{2}-\frac{1}{5}\)
\(=\frac{3}{10}\)
\(b.\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}\)
\(=2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}\right)\)
\(=2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)\)
\(=2\cdot\left(\frac{1}{2}-\frac{1}{5}\right)\)
\(=2\cdot\frac{3}{10}=\frac{3}{5}\)
\(c.\frac{1}{2\cdot3}+\frac{2}{3\cdot5}+\frac{3}{5\cdot8}\)
\(=\frac{1}{6}+\frac{2}{15}+\frac{3}{40}\)
\(=\frac{3}{8}\)
k nha 500 AE
a, \(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}\)
\(=\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)
\(=\frac{1}{2}-\frac{1}{5}\)
\(=\frac{3}{10}\)
b, \(\frac{2}{2\times3}+\frac{2}{3\times4}+\frac{2}{4\times5}\)
\(=\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}\)
\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)\times\frac{2}{1}\)
\(=\left(\frac{1}{2}-\frac{1}{5}\right)\times\frac{2}{1}\)
\(=\frac{3}{10}\times\frac{2}{1}\)
\(=\frac{3}{5}\)
c, \(\frac{1}{2\times3}+\frac{2}{3\times5}+\frac{3}{5\times8}\)
\(=\frac{3-2}{2\times3}+\frac{5-3}{3\times5}+\frac{8-5}{5\times8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}\)
\(=\frac{3}{8}\)
Nhận xét : \(\frac{1}{6}=\frac{1}{2\cdot3}\);\(\frac{1}{12}=\frac{1}{3\cdot4}\);\(\frac{1}{20}=\frac{1}{4\cdot5};...\)
5 phân số còn lại : \(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}\)
Tổng là :
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{11\cdot12}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{2}-\frac{1}{12}\)
\(=\frac{5}{12}\)
ta có:
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...\)
Từ đây ta có thể suy luận ra 5 p/số sau
Ta có :
\(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{13.14}\)
\(C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{13}-\frac{1}{14}\)
\(C=\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(1-\frac{1}{14}\right)\)
\(C=1-\frac{1}{14}\)
\(C=\frac{14}{14}-\frac{1}{14}\)
\(C=\frac{14-1}{14}\)
\(C=\frac{13}{14}\)
Vậy \(C=\frac{13}{14}\)
Chúc bạn học tốt ~
\(C=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{13\cdot14}\)
\(C=\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+\frac{4-3}{3\cdot4}+....+\frac{14-13}{13\cdot14}\)
\(C=\frac{2}{1\cdot2}-\frac{1}{1\cdot2}+\frac{3}{2\cdot3}-\frac{2}{2\cdot3}+\frac{4}{3\cdot4}-\frac{3}{3\cdot4}+....+\frac{14}{13\cdot14}-\frac{13}{13\cdot14}\)
\(C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{13}-\frac{1}{14}\)
\(C=1-\frac{1}{14}\)
\(C=\frac{13}{14}\)
dấu "." là dấu nhân nhs
1/3-1/4 + 1/4-1/5 + 1/5-1/6 + 1/6-1/7 +...+ 1/11-1/12= 1/3-1/12 =1/4
\(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{11\cdot12}\)
\(=\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+...+\left(\frac{1}{11}-\frac{1}{12}\right)\)
\(=\frac{1}{3}-\frac{1}{12}=\frac{1}{4}\)