a) So sánh \(1995^n.1997^n\)với \(1996^{2n}\)
b) Rút gọn \(A=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+\frac{1}{12.15}+\frac{1}{15.18}+\frac{1}{18.21}+\frac{1}{21.24}\)
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23 tháng 8 2016
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 \(B=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+\frac{1}{12.15}\)
\(3B=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+\frac{3}{12.15}\)
\(3B=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\)
\(3B=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
\(B=\frac{4}{15}\div3=\frac{4}{45}\)
25 tháng 8 2016
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}=\frac{100}{101}\)
\(A=\frac{100}{101}\div2=\frac{50}{101}\)
15 tháng 4 2018
a,A=\(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{23.24}\)
A=\(\frac{1}{2}+\frac{2}{1}-\frac{1}{3}+\frac{3}{1}-\frac{1}{4}+......\frac{23}{1}-\frac{1}{24}\)
A=\(\frac{1}{2}-\frac{1}{24}\)
A=\(\frac{11}{24}\)
6 tháng 3 2022
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{15}-\dfrac{1}{18}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{18}\right)=\dfrac{1}{3}\cdot\dfrac{5}{18}=\dfrac{5}{54}\)
24 tháng 7 2020
Trả lời:
a, \(\frac{6\times9-2\times17}{63\times3-119}=\frac{2.3\times9-2\times17}{7.9\times3-7.17}\)
\(=\frac{2\times\left(3\times9-17\right)}{7\times\left(3\times9-17\right)}\)
\(=\frac{2}{7}\)
b, \(\frac{3\times13-13\times18}{15\times40-80}=\frac{13\times\left(3-18\right)}{40\times\left(15-2\right)}\)
\(=\frac{13\times-15}{40\times13}\)
\(=\frac{-3}{8}\)
c, \(\frac{-1997.1996+1}{\left(-1995\right).\left(-1997\right)+1996}=\frac{-1997.1996+1}{\left(1-1996\right).\left(-1997\right)+1996}\)
\(=\frac{-1997.1996+1}{-1997-1996.\left(-1997\right)+1996}\)
\(=\frac{-1997.1996+1}{-1996.\left(-1997\right)-1}\)
\(=\frac{-1997.1996+1}{-\left[1996.\left(-1997\right)+1\right]}\)
\(=-1\)
d, \(\frac{3.7.13.37.39-10101}{505050-70707}=\frac{10101.39-10101}{50.10101-7.10101}\)
\(=\frac{10101.\left(39-1\right)}{10101.\left(50-7\right)}\)
\(=\frac{10101.38}{10101.43}\)
\(=\frac{38}{43}\)
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/
\(A=\frac{1}{2.9}+\frac{1}{6.9}+\frac{1}{9.12}+\frac{1}{9.20}+\frac{1}{9.30}+\frac{1}{9.42}+\frac{1}{9.56}\)
\(A=\frac{1}{9}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\right)\)
\(A=\frac{1}{9}\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{8-7}{7.8}\right)\)
\(A=\frac{1}{9}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\right)\)
\(A=\frac{1}{9}\left(1-\frac{1}{9}\right)=\frac{1}{9}.\frac{8}{9}=\frac{8}{81}\)