tìm x biết :(1/1.2.3+1/2.3.4+1/3.4.5+..........+1/98.99.100).x=-3
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DK
0
NT
5
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
27 tháng 1 2017
Nhân cả hai vế với 2
\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}=\frac{2}{x}\left(\frac{1}{1.2}-\frac{1}{99.100}\right).\)
Xét vế trái
\(VT=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\)
\(VT=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(VT=\frac{1}{1.2}-\frac{1}{99.100}\)
\(\Rightarrow\frac{2}{x}=1\Rightarrow x=2\)
TH
0
HT
2 tháng 8 2015
1/1.2.3 + 1/2.3.4 +....+1/98.99.100
= 1/2 . (3-1/1.2.3 + 4-2/2.3.4 +....+ 100-98/98.99.100)
= 1/2 . (3/1.2.3 -1/1.2.3 + 4/2.3.4 - 2/2.3.4 +.......+ 100/98.99.100 - 98/98.99.100)
= 1/2 . (1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 +......+ 1/98.99 - 1/99.100)
= 1/2 . (1/2 - 1/9900)
= 1/2 . 4949/9900
= 4949/19800
EC
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
20 tháng 7 2023
a/
\(b=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)
\(2b=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{99-97}{97.99}=\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}=\)
\(=1-\dfrac{1}{99}=\dfrac{98}{99}\Rightarrow b=\dfrac{98}{2.99}=\dfrac{49}{99}\)
b/
\(c=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}=\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+\dfrac{1}{98.99}-\dfrac{1}{99.100}=\)
\(=\dfrac{1}{2}-\dfrac{1}{99.100}\)
c/
\(\dfrac{2}{5}.d=\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}+\dfrac{101-99}{99.100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}+\dfrac{1}{99.100}-\dfrac{1}{100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{100.101}\Rightarrow d=\left(\dfrac{1}{2.3}-\dfrac{1}{100.101}\right):\dfrac{2}{5}\)
L
2
TN
17 tháng 9 2016
B=1/1.2.3+1/2.3.4+1/3.4.5+............+1/98.99.100
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}\cdot\frac{4949}{9900}\)
\(=\frac{4949}{19800}\)
17 tháng 9 2016
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(B=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
\(B=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(B=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(B=\frac{1}{2}.\frac{4949}{9900}=\frac{4949}{19800}\)
\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\right)x=-3\)
\(\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)x=-3\)
\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{9900}\right)x=-3\)
\(\dfrac{1}{2}.\dfrac{4949}{9900}x=-3\)
\(\dfrac{4949}{19800}x=-3\)
\(x=-3:\dfrac{4949}{19800}\)
\(x=-\dfrac{59400}{4949}\)
thank you very good