G=\(\frac{1}{3}\)+\(\frac{1}{15}\)+.....+\(\frac{1}{9999}\)
Những câu hỏi liên quan
\(1005+\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9603}+\frac{1}{9999}\)
1005\(\frac{49}{303}\)
A(-1,7).2,3+1,7.(-3,7)-1,7.3-0,17:frac{1}{10}Bleft(frac{1}{2}-1right).left(frac{1}{3}-1right).left(frac{1}{4}-1right)...left(frac{1}{2017}-1right)Cfrac{3}{4}.frac{8}{9}.frac{15}{16}.....frac{899}{900}Dfrac{1}{3}+frac{1}{15}+frac{1}{35}+...+frac{1}{9999}E1-3+3^2-3^3+...+3^{2016}-3^{2017}+3^{2018}G2+2^2+2^3+...+2^{60}
Đọc tiếp
A=(-1,7).2,3+1,7.(-3,7)-1,7.3-0,17:\(\frac{1}{10}\)
B=\(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).\left(\frac{1}{4}-1\right)...\left(\frac{1}{2017}-1\right)\)
C=\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{899}{900}\)
D=\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9999}\)
E=\(1-3+3^2-3^3+...+3^{2016}-3^{2017}+3^{2018}\)
G=\(2+2^2+2^3+...+2^{60}\)
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{99.101}\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(A=\frac{1}{2}.\frac{98}{303}\)
\(A=\frac{49}{303}\)
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A= \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
2A=\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
2A=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
2A=\(\frac{1}{3}-\frac{1}{101}\)
2A=\(\frac{98}{303}\)
A=\(\frac{98}{303}.\frac{1}{2}\)
A=\(\frac{49}{303}\)
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Tính giá trị của biểu thức, tính nhanh nếu có thể:
21) ( 1 + 1/3 ) . ( 1 + 1/8 ) . ( 1 + 1/15 ) . ...... ( 1+ 1/9999)
22) A = \(\frac{38}{25}+\frac{9}{10}-\frac{11}{15}+\frac{13}{21}-\frac{15}{28}+\frac{17}{36}-......+\frac{197}{4851}-\frac{199}{4950}\)
21)
\(\left(1+\dfrac{1}{3}\right).\left(1+\dfrac{1}{8}\right).\left(1+\dfrac{1}{15}\right).....\left(1+\dfrac{1}{9999}\right)\\ =\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.....\dfrac{10000}{9999}\\ =\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{100.100}{99.101}\\ =\dfrac{2.3.4.....100}{1.2.3.....99}.\dfrac{2.3.4.....100}{3.4.5.....101}\\ =100.\dfrac{2}{101}\\ =\dfrac{200}{101}\)
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\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}=?\)
TÍNH:
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9999}\)
GIÚP MÌNH VỚI, MIK SẼ TICK CHO!!!
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+....+\frac{1}{9999}\)
=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{99.101}\)
=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
=\(1-\frac{1}{101}=\frac{100}{101}\)
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\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9999}\)
\(2\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{9999}\right)\)
\(2\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(2\left(1-\frac{1}{101}\right)\)
\(2\times\frac{100}{101}\)
\(\frac{200}{101}\)
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Xem thêm câu trả lời
A = \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+.....+\frac{1}{9999}\)
A = ?
\(\Rightarrow A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{99.101}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{88}{303}\)
\(\Rightarrow A=\frac{44}{303}\)
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\(A=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{99\times101}\)
\(\Rightarrow2A=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{99\times101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
=> A = 98/203 : 2 = 49/303
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\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+...+\frac{1}{99\cdot101}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\)
\(A=\frac{1}{3}-\frac{1}{101}\)
\(A=\frac{98}{303}\)
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Xem thêm câu trả lời
Tính nhanh : \(\frac{10000}{10001}-\frac{9999}{10000}+\frac{1}{9999}-\frac{1}{10000}+...+\frac{3}{4}-\frac{2}{3}+\frac{1}{2}-\frac{1}{3}\)
Tính nhanh: \(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)
A=1/3.5+1/5.7+1/7.9+...+1/99.101
2A= 2/3.5+2/5.7+2/7.9+...+2/99.101
2A= 1/3-1/5+1/5-1/7-1/7+1/7-1/9+...+1/99-1/101
2A=1/3-1/101=98/303
A=(98/303)/2=49/303
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A=1/3.5 + 1/5.7 + 1/7.9 +...+ 1/99.101
=1/2.[(1/3-1/5) + (1/5-1/7) + ... + 1/99-1/101)]
=1/2.(1/3-1/101)
=49/303
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