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có thể đây là bài lớp 4 nhưng mình nghĩ là các bạn lớp 5 cũng sẽ khó khăn đó
A = \(\frac{24}{48}\)+ \(\frac{12}{48}\)+ \(\frac{8}{48}\)+ \(\frac{2}{48}\)+ \(\frac{1}{48}\)
A = \(\frac{24+12+8+2+1}{48}\)= \(\frac{47}{48}\)
ai tốt bụng thì tk cho mk nha
\(\frac{5}{6}\)\(+\frac{41}{6}+\left(\frac{225}{20}-\frac{37}{4}\right):\frac{25}{3}=\frac{23}{3}+2:\frac{25}{3}=\frac{23}{3}+\frac{6}{25}=\frac{593}{75}\)
\(\frac{5}{6}+6\frac{5}{6}.\left(11\frac{5}{20}-9\frac{1}{4}\right):8\frac{1}{3}\)
\(=\frac{5}{6}+\frac{41}{6}.\left(\frac{45}{4}-\frac{37}{4}\right):\frac{25}{3}\)
\(=\frac{5}{6}+\frac{41}{6}.2.\frac{3}{25}\)
\(=\frac{5}{6}+\frac{41}{25}\)
\(=\frac{371}{150}\)
đơn giản :
A=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+........+\(\frac{1}{99.100}\)
A= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
A=1 - \(\frac{1}{100}\)
A= \(\frac{99}{100}\)
CÓ AI DÙNG HỌC 24 GIỜ KO
A = 1/2 + 1/6 / + 1/ 12 + 1/20 + ......+ 1/(99.100)
A= 1/ ( 1 x 2 ) + 1/ ( 2 x 3 ) + 1 / ( 3 x 4 ) + .....+ 1/ ( 99 x 100 )
A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + .................+ 1/99 - 1/100
A= 1 - 1/100
A= 99/100
CHÚC BẠN HỌC TỐT
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+...+\frac{1}{72}+\frac{1}{81}\)
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{8\times9}+\frac{1}{81}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{81}\)
\(A=1-\frac{1}{9}+\frac{1}{81}=\frac{73}{81}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{81}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{81}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{81}\)
\(=1-\frac{1}{9}+\frac{1}{81}\)
\(=\frac{8}{9}+\frac{1}{81}\)
\(=\frac{73}{81}\)
\(\frac{1}{2}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{20}\)+ \(\frac{1}{30}\)+ \(\frac{1}{42}\)+ \(\frac{1}{56}\)
= \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ ...... + \(\frac{1}{7.8}\)
= \(1\)\(-\)\(\frac{1}{8}\)
= \(\frac{7}{8}\)
thiếu bước :v
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)
\(=1-\frac{1}{8}\)
\(=\frac{7}{8}\)
a) \(\left(\frac{4}{3}-\frac{4}{6}\right)+\left(\frac{4}{6}-\frac{4}{9}\right)+\left(\frac{4}{9}-\frac{4}{10}\right)+\left(\frac{4}{12}-\frac{4}{15}\right)\)
\(=\frac{4}{15}-\frac{4}{3}=\frac{-16}{15}\)
C) bạn chỉ ần bỏ các số giống nhau thôi nhé
= 1
b)
\(C=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
\(C=1\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{3}+\frac{1}{3}\times\frac{1}{4}+\frac{1}{4}\times\frac{1}{5}+\frac{1}{5}\times\frac{1}{6}\)
\(C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(C=1-\frac{1}{6}\)
\(C=\frac{5}{6}\)
\(=\frac{5}{6}nhé\)