Tính
S=\(\frac{1}{1.3}\)- \(\frac{1}{24}\) + \(\frac{1}{3.5}\)-\(\frac{1}{4.6}\)+...................+ \(\frac{1}{97.99}\)- \(\frac{1}{98.100}\)
( GIÚP MIK VS MIK ĐANG CẦN GẤP LẮM Ạ ! )
Chứng minh rằng: \(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+...+\frac{1}{97.99}+\frac{1}{98.100}< \frac{3}{4}\)
Tính tổng: \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
giúp mik vs, ai nhanh tick luôn nè:)))
=49/99 NHA
HT
k cho mình nha
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\(P=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)
2A=\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)
2A=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\)
2A=\(1-\frac{1}{99}\)
A=\(\frac{49}{99}\)
Chúc bạn học tốt
HYC-30/1/2022
Tính \(\frac{1}{3.1}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+...+\frac{1}{97.99}-\frac{1}{98.100}\)
\(\frac{1}{3.1}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+...+\frac{1}{97.99}-\frac{1}{98.100}\)
= \(\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
= \(\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{100}\right)\)
= \(\frac{1}{2}\left(1-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}.\frac{98}{99}-\frac{1}{2}.\frac{49}{100}\)
= \(\frac{49}{99}-\frac{49}{200}\)
= \(\frac{4949}{19800}\)
bn zô xem nha, ko hiểu thì cứ hỏi bn ấy nhá
http://olm.vn/hoi-dap/question/154321.html
A=\(\frac{1}{3.1}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+...+\frac{1}{97.99}-\frac{1}{98.100}=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
Ta có:\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)=\frac{1}{2}.\left(1-\frac{1}{99}\right)=\frac{1}{2}\cdot\frac{98}{99}=\frac{49}{99}\)
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}\cdot\frac{49}{100}=\frac{49}{200}\)
=>A=49/99-49/200=4949/19800
Tính nhanh
1,
A = \(\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+..........+\frac{1}{97.99}-\frac{1}{98.100}\)
2,
B=
\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+.........+\frac{1}{20}.\left(1+2+3+.........+20\right)\)
Thực hiện phép tính:
a)\(\frac{1}{6}-\frac{2}{3}\)
b)\(\frac{2.3+4.6+14.21+18.27+26.39}{3.5+6.10+21.35+27.45+39.65}\)
CÁC BẠN GIÚP MIK VS, MIK ĐANG CẦN GẤP!!!
\(a,\frac{1}{6}-\frac{2}{3}\)
\(=\frac{1}{6}-\frac{4}{6}=-\frac{3}{6}=-\frac{1}{2}\)
a, -1/2
b, 6517/9805
\(b,\frac{2.3+4.6+14.21+18.27+26.39}{3.5+6.10+21.35+27.45+39.65}\)
\(=\frac{2+4+14+18+26}{5+10+35+45+65}\)
\(=\frac{64}{115}\) ( ko chắc )
A= \(\frac{1}{2.4}-\frac{1}{3.5}+\frac{1}{4.6}-\frac{1}{5.7}+.....-\frac{1}{97.99}+\frac{1}{96.98}-\frac{1}{97.99}\)
\(A=\left(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+.........+\frac{1}{96\cdot98}\right)-\left(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+.......+\frac{1}{97\cdot99}\right)\)
\(=\frac{1}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+....+\frac{2}{96\cdot98}\right)-\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+.....+\frac{2}{97\cdot99}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{96}-\frac{1}{98}\right)-\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{98}\right)-\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=\frac{12}{49}-\frac{16}{99}=\frac{404}{4851}\)
Tính: A=(1-\(\frac{1}{1.3}\)).(1-\(\frac{1}{2.4}\)).(1-\(\frac{1}{3.5}\))...(1-\(\frac{1}{98.100}\))
Giúp mk nhé mn. Mk đang cần gấp.Nhớ giải chi tiết nhé. Mk tick cho.
a) (\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{132}\)) . x =\(\frac{1}{3}\)
b) (\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)) : x = \(\frac{2}{3}\)
c) (\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)) . x = \(\frac{2}{3}\)
Mik đang cần gấp
a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)
(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)
(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)
x=\(\frac{1}{3}:\frac{13}{12}\)
x=\(\frac{4}{13}\)
b)( \(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+...+\frac{2}{9}-\frac{2}{11}_{ }\)):x =\(\frac{2}{3}\)
Giống câu a
\(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{97.99}\)
2A=2-1/3+1/3-1/5+...+1/97-1/99
2A=2-1/99
2A=197/99
A=197/198
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{97\cdot99}\)
\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{97\cdot99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}\cdot\frac{98}{99}\)
\(=\frac{49}{99}\)
=))
\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\frac{98}{99}\)
\(=\frac{49}{99}\)