1/3.6+1/6.9+1/9.12+1/12.15+1/15.18
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/
\(A=\frac{1}{2.9}+\frac{1}{6.9}+\frac{1}{9.12}+\frac{1}{9.20}+\frac{1}{9.30}+\frac{1}{9.42}+\frac{1}{9.56}\)
\(A=\frac{1}{9}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\right)\)
\(A=\frac{1}{9}\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{8-7}{7.8}\right)\)
\(A=\frac{1}{9}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\right)\)
\(A=\frac{1}{9}\left(1-\frac{1}{9}\right)=\frac{1}{9}.\frac{8}{9}=\frac{8}{81}\)
\(\left(\frac{-1}{3.6}-\frac{1}{6.9}-\frac{1}{9.12}-\frac{1}{12.15}:\left|x\right|=\frac{-8}{15}\right)\)
4/3.6+ 4/6.9+ 4/9.12 + 4/12.15
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 \(B=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+\frac{1}{12.15}\)
\(3B=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+\frac{3}{12.15}\)
\(3B=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\)
\(3B=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
\(B=\frac{4}{15}\div3=\frac{4}{45}\)
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}=\frac{100}{101}\)
\(A=\frac{100}{101}\div2=\frac{50}{101}\)
\(\frac{4}{3.6}\)+ \(\frac{4}{6.9}\)+ \(\frac{4}{9.12}\)+\(\frac{4}{12.15}\)
= 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
Chứng minh: 1/3.6 + 1/6.9 + 1/9.12 + ... + 1/219.222 <1
\(S=\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{219.222}\)
\(\Rightarrow3S=\frac{3}{3.6}+\frac{3}{6.9}+...+\frac{3}{219.222}\)
\(=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{219}-\frac{1}{222}\)
\(=\frac{1}{3}-\frac{1}{222}< \frac{1}{3}\)
\(\Rightarrow S< \frac{1}{9}< 1\)
\(\Rightarrow S< 1\left(đpcm\right)\)
số hạng của dãy: 1/3.6 ; 1/6.9 ; 1/9.12 ; ... ; 1/156.159
Số số hạng là :
(159 - 6) : 3 + 1 = 52 (số hạng)
Số số hạng là :
(159 - 6) : 3 + 1 = 52 (số hạng)
Số số hạng của dãy 1/3.6; 1/6.9; 1/9.12;...; 1/156.159
Số số hạng là:(159-6):3+1=52(số hạng)
số số hạng của dãy 1/3.6 + 1/6.9 + 1/9.12 + ... + 1/156.159 là...
Số số hạng là:(159-6):3+1=52(số hạng)