Tìm A
A=\(\frac{1}{6}\)+ \(\frac{1}{12}\)+\(\frac{1}{20}\)+...+\(\frac{1}{9900}\)
Chúc các bạn làm tốt bài này và hãy giúp mình nhiều bài nữa nhé!
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Những câu hỏi liên quan
31 tháng 5 2018
\(\frac{1}{2}+\frac{1}{6}\)\(+\frac{1}{12}\)\(+...+\frac{1}{9702}\)\(+\frac{1}{9900}\)
= \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}\)\(+...+\frac{1}{98\cdot99}\)+ \(\frac{1}{99\cdot100}\)
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\)\(\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
= \(\frac{1}{1}-\frac{1}{100}\)
= \(\frac{100}{100}\)- \(\frac{1}{100}\)
= \(\frac{99}{100}\)
31 tháng 5 2018
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9702}+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
6 tháng 3 2016
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(A=\frac{1}{2}-\frac{1}{8}\)
\(A=\frac{3}{8}\)
6 tháng 3 2016
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(A=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
mình nhé!
28 tháng 6 2020
Bài làm
\(D=\frac{6}{3,5}+\frac{6}{5.7}+...+\frac{6}{21.23}\)
\(D=3.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{21.23}\right)\)
\(D=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{21}-\frac{1}{23}\right)\)
\(D=3.\left(\frac{1}{3}-\frac{1}{23}\right)\)
\(D=3.\frac{20}{69}\)
\(D=\frac{20}{23}\)
Học tốt
28 tháng 6 2020
Bài làm
\(D=\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{21.23}\)
\(D=3.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{21.23}\right)\)
\(D=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{21}-\frac{1}{23}\right)\)
\(D=3.\left(\frac{1}{3}-\frac{1}{23}\right)\)
\(D=3.\frac{20}{69}\)
\(D=\frac{20}{23}\)
\(E=\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\)
\(E=10.\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)\)
\(E=10.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\)
\(E=10.\left(\frac{1}{11}-\frac{1}{55}\right)\)
\(E=10.\frac{4}{55}\)
\(E=\frac{8}{11}\)
\(G=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)
\(G=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(G=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(G=\frac{1}{1}-\frac{1}{100}\)
\(G=\frac{99}{100}\)
Nhớ k cho m nha
L
3 tháng 3 2017
a, A lớn nhất khi 7x la nguyên dương nho nhất
\(\Rightarrow7x=1\)
\(\Rightarrow x=\frac{1}{7}\)
\(b,B=\frac{10+4-x}{4-x}\)
\(B=\frac{10}{4-x}+1\)
b lon nhat khi 4-xla nguyen duong nho nhat
\(\Rightarrow4-x=1\)
\(\Rightarrow x=4-1=3\)
\(c,C=\frac{27-2x}{12-x}=\frac{3+24-2x}{12-x}=\frac{3}{12-x}+2\)
c lon nhat khi 12-x la nguyen duong nho nhat
\(\Rightarrow12-x=1\Rightarrow x=11\)
Ta có :
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}\)
\(A=\frac{49}{100}\)
Chúc bạn học tốt ~
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(\Leftrightarrow A=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(\Leftrightarrow A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow A=\frac{1}{2}-\frac{1}{100}\)
\(\Leftrightarrow A=\frac{49}{100}\)
Vậy A=\(\frac{49}{100}\)