\(B=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{...2}{9999}=?\)
Những câu hỏi liên quan
Tính E=\(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+...+\frac{9998}{9999}\)
\(E=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+...+\left(1-\frac{1}{9999}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{9999}\right)\)
\(=50-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)
\(=50-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=50-\frac{1}{2}.\left(1-\frac{1}{101}\right)=50-\frac{1}{2}.\frac{100}{101}=50-\frac{50}{101}=\frac{5000}{101}\)
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Tính nhanh:
\(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+...+\frac{9998}{9999}\)
\(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+...+\frac{9998}{9999}\)
\(=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+\left(1-\frac{1}{35}\right)+\left(1-\frac{1}{63}\right)+...+\left(1-\frac{1}{9999}\right)\)
\(=\left(1-\frac{1}{1\cdot3}\right)+\left(1-\frac{1}{3\cdot5}\right)+\left(1-\frac{1}{5\cdot7}\right)+\left(1-\frac{1}{7\cdot9}\right)+...+\left(1-\frac{1}{99\cdot101}\right)\)
\(=\left(1+1+1+1+...+1\right)-\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
Có tất cả : (101 - 3) : 2 + 1 = 50 chữ số 1 => (1 + 1 + 1 + .... + 1) = 1 x 50 = 50
\(\Rightarrow50-\frac{1}{2}\cdot\left(1-\frac{1}{101}\right)\)
\(=50-\frac{1}{2}\cdot\frac{100}{101}=50-\frac{100}{101}=\frac{4950}{101}\)
Vậy \(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+...+\frac{9998}{9999}=\frac{4950}{101}\)
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\(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{9998}{9999}\)
Tính \(M=\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+...+\frac{9998}{9999}\)
\(A=\frac{3}{2}+\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+...+\frac{13}{53}\)\(B=\frac{16}{15}+\frac{36}{35}+\frac{63}{63}+...+\frac{10000}{9999}\)
\(Giải-giúp-mk-nha\)
thực hiện phép tính(tính hợp lí nếu có thể)
\(\left(\frac{2^5}{3}+\frac{2^5}{15}+\frac{2^5}{35}+...+\frac{2^5}{9999}\right)\)
làm gấp cho mik với(cảm ơn)
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=\(2+2^2+2^3+...+2^{60}\)
Bài 1: Tính giá trị biểu thức:
Afrac{4}{3}+frac{4}{15}+frac{4}{35}+...+frac{4}{9999}
Bài 2: Cho Bfrac{1}{2^2}+frac{1}{3^2}+...+frac{1}{9^2}
Chứng tỏ: frac{2}{5} Bfrac{8}{9}
Bài 3: Tính giá trị biểu thức:
Cfrac{1}{2}-frac{1}{6}-frac{1}{12}-frac{1}{20}-...-frac{1}{9900}
Đọc tiếp
Bài 1: Tính giá trị biểu thức:
\(A=\frac{4}{3}+\frac{4}{15}+\frac{4}{35}+...+\frac{4}{9999}\)
Bài 2: Cho \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}\)
Chứng tỏ: \(\frac{2}{5}< B=\frac{8}{9}\)
Bài 3: Tính giá trị biểu thức:
\(C=\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-...-\frac{1}{9900}\)
Rút gọn
\(\frac{3-\frac{1}{5}+\frac{3}{20}}{2+\frac{1}{4}-\frac{3}{5}}\)
Tính
a/ \(\frac{3}{4}.\frac{8}{9}.\frac{15}{10}....\frac{9999}{1000}\)
b/ \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1\)
Tính
a)
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\\ =\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}....\frac{99.101}{100}\\ \)
\(=\left(\frac{1.2.3...99}{2.3...100}\right).\left(\frac{3.4.5...101}{2.3.4...100}\right)\\ =\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
b)
\(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{n^2}\\ < \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\\ \)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{n-1}-\frac{1}{n}\\ =1-\frac{1}{n}< 1\)
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