Tính:
\frac{1^2}{1.2} . \frac{2^2}{2.3} ... \frac{17^2}{17.18}.\frac{18^2}{18.19}=1.212.2.322...17.18172.18.19182
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Đặt \(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{18.19}+\frac{2}{19.20}\)
\(A=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(A=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(A=2\left(1-\frac{1}{20}\right)\)
\(A=2.\frac{19}{20}=\frac{19}{10}\)
Vậy ...
=2.(\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+......+\(\frac{1}{19.20}\))
=2.( 1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+..........+\(\frac{1}{19}\)-\(\frac{1}{20}\))
=2.(1-\(\frac{1}{20}\))
=2.\(\frac{19}{20}\)
= \(\frac{19}{10}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)
\(A=1-\frac{1}{20}\)
\(A=\frac{19}{20}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)
\(=1-\frac{1}{20}\)
\(=\frac{19}{20}\)
Ta có:A: 1/1.2 +1/2.3 +1/3.4+...+1/18.19+1/19.20
=> A= 1-1/2 +1/2-1/3+1/3-1/4+...+1/18-1/19+1/19-1/20
=>A= 1-1/20=19/20
Lời giải:
$A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+....+\frac{19-18}{18.19}+\frac{20-19}{19.20}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}$
$=1-\frac{1}{20}=\frac{19}{20}$
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\)
=1-1/20
=19/20
Đặt biểu thức trên là A
Ta có: A =(1^2 . 2^2 . 3^2 . 4^2)/(1.2.2.3.3.4.4.5)
= [(1.2.3.4).(1.2.3.4)] / [(1.2.3.4).(2.3.4.5)]
= 1/5
Vậy A = 1/5
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)
=\(\frac{1.1.2.2.3.3.4.4}{1.2.2.3.3.4.4.5}\)
=\(\frac{1}{5}\)