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Ta có: 2015/501=4+11/501 =>a=4
501/11=45+6/11 =>b=45
11/6=3+2/3 =>c=3
3/2=1+1/2 =>d=1
2/1=2 =>e=2
Vậy a=4 :b=45 :c=3 :d=1: e=2
Chúc bạn học tốt . Để dễ hiểu bạn hãy tham hảo đề toán giải máytính cầm tay
chung to :C = \(\frac{1}{1.1!}+\frac{1}{2.2!}+\frac{1}{3.3!}+...+\frac{1}{2019.2019!}< \frac{3}{2}\)
Thấy : \(\frac{1}{1.1!}=\frac{1}{1}\)
\(\frac{1}{2.2!}=\frac{1}{4}\)
\(\frac{1}{3.3!}< \frac{1}{1.2.3}\)( Vì 3.3! > 1.2.3 )
...
\(\frac{1}{2019.2019!}< \frac{1}{2017.2018.2019}\)( vì 2019.2019! < 2017.2018.2019)
Cộng từng vế có :
\(\frac{1}{3.3!}+\frac{1}{4.4!}+...+\frac{1}{2019.2019!}< \frac{1}{1.2.3}+...+\frac{1}{2017.2018.2019}\)
\(\Rightarrow\frac{1}{1.1!}+\frac{1}{2.2!}+...+\frac{1}{2019.2019!}< \frac{1}{1}+\frac{1}{4}+\frac{1}{1.2.3}+...+\frac{1}{2017.2018.2019}\)
\(\Rightarrow C< \frac{1}{1}+\frac{1}{4}+\left(\frac{1}{1.2}-\frac{1}{2.3}+...+\frac{1}{2017.2018}-\frac{1}{2018.2019}\right):2\)
\(\Rightarrow C< \frac{1}{1}+\frac{1}{4}+\left(\frac{1}{2}-\frac{1}{2018.2019}\right):2\)
\(\Rightarrow C< \frac{3}{2}-\frac{1}{2.2018.2019}\)
Vì \(\frac{1}{2.2018.2019}>0\Rightarrow C< \frac{3}{2}\)
b./ \(\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)(b)
Mà \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}< 0\)
(b) \(\Leftrightarrow x+2010=0\Leftrightarrow x=-2010\)
a./
\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0.\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)(a)
Mà \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}>0\)
(a) \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
S= 1/2 - 1/2 + 1/3 - 1/3 + 1/4 - 1/4 +...+ 1/50 - 1/50
S= 0 + 0 + 0 +...+ 0
S= 0
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+.....+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+.....+\frac{19}{81.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+....+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
\(=1-\frac{1}{10^2}< 1\)
Cho \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}\)... là A, ta có:
A = \(\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
A = \(\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{2^2}+...\frac{1}{9^2}-\frac{1}{10^2}\)
A = 1 \(-\frac{1}{10^2}\) <1
Vậy: A < 1
Đặt A = \(\frac{1}{2}+\frac{1}{2.2}+\frac{1}{2.2.2}+....+\frac{1}{2.2.2.2.2.2.2.2.2.2}\)
=> A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{10}}\)
=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^9}\)
=> 2A - A = \(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^9}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
=> A = \(1-\frac{1}{2^{10}}\)