\(F=\frac{2}{3.6}+\frac{2}{6.9}+\frac{2}{9.12}+...+\frac{2}{30.33}+\frac{3}{33.36}\)=?
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Tính:
\(\frac{2}{3.6}+\frac{2}{6.9}+\frac{2}{9.12}+...+\frac{2}{117.120}\)
LƯU Ý: Dấu "." là nhân nhé !
Thực hiên phép tính: \(D=\frac{1}{120}-\frac{2}{30.33}-\frac{2}{33.36}-...-\frac{2}{117.120}\)
\(A=\frac{5}{3.6}+\frac{5}{6.9}+\frac{5}{9.12}+..........+\frac{5}{96.99}\)
3/5 A = 3/3.6 + 3/6.9 +..... + 3/96.99
= 1/3 - 1/6 + 1/6 - 1/9 + .... + 1/96 - 1/99 = 1/3 - 1/99 = 32/99
=> A = 160/297
k mk nha
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\(\left(\frac{-1}{3.6}-\frac{1}{6.9}-\frac{1}{9.12}-\frac{1}{12.15}:\left|x\right|=\frac{-8}{15}\right)\)
số số hạng của dãy\(\frac{1}{3.6};\frac{1}{6.9};\frac{1}{9.12};...;\frac{1}{156.159}\)là
số số hạng của dãy \(\frac{1}{3.6};\frac{1}{6.9};\frac{1}{9.12};...;\frac{1}{156.159}\) là
số hạng của dãy trên là 52 vì (159 - 3) : 3 + 1= 52
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frac{4}{3.6}+ frac{4}{6.9}+ frac{4}{9.12}+frac{4}{12.15} 4(frac{1}{3.6}+frac{1}{6.9}+frac{1}{9.12}+frac{1}{12.15})frac{4}{3}( frac{1}{3}-frac{1}{6}+frac{1}{6}-frac{1}{9}+frac{1}{9}-frac{1}{12}+frac{1}{12}-frac{1}{15} )frac{4}{3}(frac{1}{3}-frac{1}{15})frac{4}{3}.frac{4}{15}frac{16}{45}mk làm đúng chưa
Đọc tiếp
\(\frac{4}{3.6}\)+ \(\frac{4}{6.9}\)+ \(\frac{4}{9.12}\)+\(\frac{4}{12.15}\)
= 4(\(\frac{1}{3.6}\)+\(\frac{1}{6.9}\)+\(\frac{1}{9.12}\)+\(\frac{1}{12.15}\))
=\(\frac{4}{3}\)( \(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\) )
=\(\frac{4}{3}\)(\(\frac{1}{3}-\frac{1}{15}\))
=\(\frac{4}{3}\).\(\frac{4}{15}\)
=\(\frac{16}{45}\)
mk làm đúng chưa
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Tính Nhanh : a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{99.101}\)
b) \(\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+\frac{1}{12.15}\)
a) Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}=\frac{100}{101}\)
\(A=\frac{100}{101}\div2=\frac{50}{101}\)
b) Đặt \(B=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+\frac{1}{12.15}\)
\(3B=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+\frac{3}{12.15}\)
\(3B=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\)
\(3B=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
\(B=\frac{4}{15}\div3=\frac{4}{45}\)
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Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}=\frac{100}{101}\)
\(A=\frac{100}{101}\div2=\frac{50}{101}\)
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a) So sánh \(1995^n.1997^n\)với \(1996^{2n}\)
b) Rút gọn \(A=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+\frac{1}{12.15}+\frac{1}{15.18}+\frac{1}{18.21}+\frac{1}{21.24}\)
a/
\(1995^n.1997^n=\left(1995.1997\right)^n\)
\(1996^{2n}=\left(1996^2\right)^n\)
\(1995.1997=\left(1996-1\right).\left(1996+1\right)=1996^2-1\)
\(\Rightarrow1995.1997< 1996^2\Rightarrow1995^n.1997^n< 1996^{2n}\)
b/
\(A=\frac{1}{2.9}+\frac{1}{6.9}+\frac{1}{9.12}+\frac{1}{9.20}+\frac{1}{9.30}+\frac{1}{9.42}+\frac{1}{9.56}\)
\(A=\frac{1}{9}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\right)\)
\(A=\frac{1}{9}\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{8-7}{7.8}\right)\)
\(A=\frac{1}{9}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\right)\)
\(A=\frac{1}{9}\left(1-\frac{1}{9}\right)=\frac{1}{9}.\frac{8}{9}=\frac{8}{81}\)
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