\(\frac{5}{2\cdot4}+\frac{5}{4\cdot6}+\frac{5}{6\cdot8}+.......+\frac{5}{100\cdot102}\)
help me , please everyone
1, Tính
a, \(\frac{5}{2\cdot4}\)+\(\frac{5}{4\cdot6}\)+\(\frac{5}{6\cdot8}\)+.......+\(\frac{5}{48\cdot50}\)
HELP ME!!!!!!!!
Đặt A
Ta có công thức :
\(\frac{a}{b.c}=\frac{a}{c-b}.\left(\frac{1}{b}-\frac{1}{c}\right)\)
Dựa vào công thức, ta có
\(A=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{48}-\frac{1}{50}\right)\)
\(A=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)=\frac{5}{2}.\left(\frac{12}{25}\right)=\frac{6}{5}\)
Ai thấy đúng thì ủng hộ nha !!!
a, \(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{48.50}\)
=\(\frac{5}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}\right)\)
=\(\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
=\(\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)=\(\frac{5}{2}.\frac{12}{25}\)=\(\frac{6}{5}\)
B=\(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{96\cdot98}+\frac{1}{98\cdot100}\)
Please help me!!!!!!!!
\(B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{96}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}.\frac{49}{100}=\frac{49}{200}\)
Bn Nguyễn Tuấn Minh làm đúng rồi đó bạn
\(B=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{96.98}+\frac{1}{98.100}\)
\(B=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{96.98}+\frac{2}{98.100}\right)\)
\(B=\frac{1}{2}.(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{96}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100})\)
\(B=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(B=\frac{1}{2}.\frac{49}{100}\)
\(B=\frac{49}{200}\)
B=( \(\frac{5}{2\cdot4}\)+ \(\frac{5}{4\cdot6}\)+ ....... +\(\frac{5}{100\cdot102}\))
\(B=\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{100.102}\)
\(B=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{5}{2}.\left(\frac{1}{4}-\frac{1}{6}\right)+...+\frac{5}{2}.\left(\frac{1}{100}-\frac{1}{102}\right)\)
\(B=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{102}\right)\)
\(B=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{102}\right)\)
\(B=\frac{5}{2}.\frac{25}{51}\)
\(B=\frac{125}{102}\)
đè là gì vậy cho mỗi phép tính để đấy à coi để người khác nghĩ đề à
Tính (theo mẫu)
Mẫu:\(\frac{5\cdot6\cdot7\cdot9}{12\cdot7\cdot27}\)=\(\frac{5\cdot6\cdot7\cdot9}{6\cdot2\cdot7\cdot9\cdot3}\)=\(\frac{5}{6}\)
a.\(\frac{3\cdot4\cdot7}{12\cdot8\cdot9}\)=.........................................................................
b.\(\frac{4\cdot5\cdot6}{12\cdot10\cdot8}\)=.......................................................................
c\(\frac{5\cdot6\cdot7}{12\cdot14\cdot15}\)=......................................................................
a.\(\frac{3\cdot4\cdot7}{12\cdot8\cdot9}\)= \(\frac{3\cdot4\cdot7}{3\cdot4\cdot8\cdot9}\)= \(\frac{7}{72}\)
b. \(\frac{4\cdot5\cdot6}{12\cdot10\cdot8}\)= \(\frac{4\cdot5\cdot2\cdot3}{3\cdot4\cdot5\cdot2\cdot8}\)= \(\frac{1}{8}\)
c.\(\frac{5\cdot6\cdot7}{12\cdot14\cdot15}\)= \(\frac{5\cdot6\cdot7}{2\cdot6\cdot2\cdot7\cdot3\cdot5}\)= \(\frac{1}{12}\)
\(20\%x+\frac{2}{5}x=x-4\)
\(\frac{\frac{2}{3}+\frac{2}{7}\cdot\frac{1}{14}}{-1-\frac{3}{7}+\frac{3}{28}}\)
\(\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20}\)
giải hội mik mới cách làm lun nha
\(\frac{7}{3\cdot4}-\frac{9}{4\cdot5}+\frac{11}{5\cdot6}-\frac{13}{6\cdot7}+\frac{15}{7\cdot8}-\frac{17}{8\cdot9}+\frac{19}{9\cdot10}\)
ta có:(3+4)/4=3/(3*4)+4/(3*4) =1/4+1/3 chứng minh tương tự,cộng vế với vế, ta được kết quả là 13/30
\(41\sqrt[9^1]{8\sqrt[2]{\frac{12}{2.85\frac{1\cdot2+3\cdot4+5\cdot6+7\cdot8+9\sqrt[4]{16}}{2\cdot\frac{12}{2}\sqrt{4^2}-7^2}}}4\cdot5\cdot6\cdot7\cdot8\cdot9}\)
Ô phép tính khủng. Cái này do bạn chế ra à !
Tính nhanh:
A=\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(A=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(A=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\\ \)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}\)
\(=\frac{4}{8}-\frac{1}{8}\\ =\frac{3}{8}\)
Chúc bn học thiệt giỏi nhé!
cảm ơn bẹn nhìu
\(\frac{1}{10\cdot9}-\frac{1}{9\cdot8}-\frac{1}{8\cdot7}-\frac{1}{7\cdot6}-\frac{1}{6\cdot5}-\frac{1}{5\cdot4}-\frac{1}{4\cdot3}-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
Ta có : \(\frac{1}{10.9}-\frac{1}{9.8}-.....-\frac{1}{2.1}\)
\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.8}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\frac{8}{9}=\frac{-79}{90}\)