chứng minh rằng \(\frac{91}{1\cdot4}+\frac{91}{4\cdot7}+\frac{91}{7\cdot10}+......+\frac{91}{88\cdot91}\)=30
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LV
2
XO
2 tháng 5 2021
Ta có \(\frac{91}{1.4}+\frac{91}{4.7}+\frac{91}{7.10}+...+\frac{91}{88.91}=\frac{91}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{88.91}\right)\)
\(=\frac{91}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{88}-\frac{1}{91}\right)\)
\(=\frac{91}{3}.\left(1-\frac{1}{91}\right)=\frac{91}{3}.\frac{90}{91}=30\)
2 tháng 5 2021
9/1.4+9/4.7+9/7.10+…+9/88.91=30
=91/3. (3/1.4+3/4.7+3/7.10+…+3/88.91)
=91/3. (1-1/4+1/4-1/7+1/7-1/10+…+1/88-1/91)
=91/3*90/91=30
RD
0
NS
0
XO
17 tháng 8 2020
a) \(\frac{91}{1.4}+\frac{91}{4.7}+...+\frac{91}{88.91}=\frac{91}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{88.91}\right)\)
\(=\frac{91}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{88}-\frac{1}{91}\right)=\frac{91}{3}\left(1-\frac{1}{91}\right)=\frac{91}{3}.\frac{90}{91}=30\left(\text{đpcm}\right)\)
29 tháng 8 2017
A=1/3.7+1/7.11+1/11.15+...+1/91.95
=>4A=4/3.7+4/7.11+4/11.15+...+4/91.95
4A=1/3-1/7+1/7-1/11+1/11-1/15+...1/91-1/95
4A=1/3-1/95
4A=92/285
A=92/285:4
A=23/285
TD
1
DT
22 tháng 4 2016
Ta có:
A=1/1.3+2/3.7+3/7.13+...+10/91.111
=>2A=2/1.3+4/3.7+6/7.13+...+20/91.111
=>2A=1-1/3+1/3-1/7+1/7-1/13+...+1/91-1/111
=>2A=1-1/111=110/111
=>A=55/111
Vậy A=55/111
OK!
8 tháng 4 2017
\(A=\frac{1}{1.4}+\frac{1}{2.7}+...+\frac{1}{67.70}\)
\(3A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{67.70}\)
\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{67}-\frac{1}{70}\)
\(3A=1-\frac{1}{70}=\frac{69}{70}\)
\(A=\frac{69}{70}:3=\frac{23}{70}\)
vì \(\frac{23}{70}< 1\)
nên \(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{67.70}< 1\)
NT
1
NT
6
#)Giải :
\(\frac{91}{1.4}+\frac{91}{4.7}+\frac{91}{7.11}+...+\frac{91}{88.91}\)
\(=\frac{91}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+...+\frac{3}{88.91}\right)\)
\(=\frac{91}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{88}-\frac{1}{91}\right)\)
\(=\frac{91}{3}\left(1-\frac{1}{91}\right)\)
\(=\frac{91}{3}.\frac{90}{91}=30\left(đpcm\right)\)
#~Will~be~Pens~#
\(\frac{91}{1\cdot4}+\frac{91}{4\cdot7}+...+\frac{91}{88\cdot91}=\frac{1}{3}\left(91-\frac{91}{4}+\frac{91}{4}-\frac{91}{7}+...-\frac{91}{91}\right)\)
\(=\frac{1}{3}\left(91-1\right)=\frac{1}{3}\cdot90=30\)