so sánh A=3/9x14+3/14x19+3/19x24+....+3/(5n-1)x(5n+4) và 1/15
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NM
0
TT
2
SY
1 tháng 4 2016
Ta có: A=\(\frac{14^{15}+3}{14^{15}+3}\) = 1
B=\(\frac{14^{16}+5}{14^{17}+5}\) < 1 => B<1=A => B<A.
DK
0
H
0
12 tháng 6 2017
\(\frac{3}{9.14}+\frac{3}{14.19}+\frac{3}{19.24}+....+\frac{3}{\left(5n+1\right)\left(5n+4\right)}\)
\(=\frac{3}{5}\left(\frac{5}{9.14}+\frac{5}{14.19}+\frac{5}{19.24}+....+\frac{5}{\left(5n+1\right)\left(5n+4\right)}\right)\)
\(=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+....+\frac{1}{5n+1}-\frac{1}{5n+4}\right)\)
\(=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{5n+4}\right)\)
\(=\frac{1}{15}-\frac{3}{5\left(5n+4\right)}< \frac{1}{15}\) (đpcm)
XO
10 tháng 2 2020
Ta có\(\frac{3}{9.14}+\frac{3}{14.19}+...+\frac{3}{\left(5n-1\right)\left(5n+4\right)}=\frac{3}{5}\left(\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{\left(5n-1\right)\left(5n+4\right)}\right)\)
\(=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{5n-1}-\frac{1}{5n+4}\right)=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{5n+4}\right)=\frac{1}{15}-\frac{3}{25n+20}\)(1)
kết hợp điều kiện ta có \(\frac{3}{25n+20}\ge\frac{3}{25.2+20}=\frac{3}{70}>0\)
=> \(\frac{3}{9.14}+\frac{3}{14.19}+...+\frac{3}{\left(5n-1\right)\left(5n+4\right)}< \frac{1}{15}\)(đpcm)
10 tháng 10 2017
Đặt :
\(A=\dfrac{3}{9.14}+\dfrac{3}{14.19}+........+\dfrac{3}{\left(5n-1\right)\left(5n+4\right)}\)
\(\Leftrightarrow\dfrac{5}{3}A=\dfrac{5}{9.14}+\dfrac{5}{14.19}+........+\dfrac{5}{\left(5n-1\right)\left(5n+4\right)}\)
\(\Leftrightarrow\dfrac{5}{3}A=\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...........+\dfrac{1}{5n-1}-\dfrac{1}{5n+4}\)
\(\Leftrightarrow\dfrac{5}{3}A=\dfrac{1}{9}-\dfrac{1}{5n+4}\)
\(\Leftrightarrow A=\left(\dfrac{1}{9}-\dfrac{1}{5n+4}\right):\dfrac{5}{3}\)
\(\Leftrightarrow A=\left(\dfrac{1}{9}-\dfrac{1}{5n+4}\right).\dfrac{3}{5}\)
\(\Leftrightarrow A=\dfrac{1}{9}.\dfrac{3}{5}-\dfrac{1}{5n+4}.\dfrac{3}{5}\)
\(\Leftrightarrow A=\dfrac{1}{15}-\dfrac{1}{5n+4}.\dfrac{3}{5}< \dfrac{1}{15}\)
\(\Leftrightarrow A< \dfrac{1}{15}\left(đpcm\right)\)
Lời giải:
$A=\frac{3}{5}\left(\frac{5}{9.14}+\frac{5}{14.19}+\frac{5}{19.24}+...+\frac{5}{(5n-1)(5n+4)}\right)$
$\frac{3}{5}\left(\frac{14-9}{9.14}+\frac{19-14}{14.19}+\frac{24-19}{19.24}+...+\frac{(5n+4)-(5n-1)}{(5n-1)(5n+4)}\right)$
$=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+\frac{1}{19}-\frac{1}{24}+...+\frac{1}{5n-1}-\frac{1}{5n+4}\right)$
$=\frac{3}{5}(\frac{1}{9}-\frac{1}{5n+4})$
$=\frac{1}{15}-\frac{3}{5(5n+4)}< \frac{1}{15}$