A=5/2x4+5/4x6+.......+5/98x100
giải theo lớp 5 nha,cám ơn nhìu
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11 tháng 6 2020
\(\frac{5}{2\cdot4}+\frac{5}{4\cdot6}+\frac{5}{6\cdot8}+.....+\frac{5}{48\cdot60}\)
\(=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+.....+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{50}\right)\)
Tự tính nốt :p
DL
3
16 tháng 8 2015
A=5/2x(2/2x4+2/4x6+2/6x8+...+2/14x16)
=5/2x(1/2-1/4+1/4-1/6+...+1/14-1/16)
=5/2x(1/2-1/16)
=5/2x(7/16)
=35/32
29 tháng 3 2016
Giải
1/2x4+1/4x6+1/6x8+...+1/96x98+1/98x100
= 1/2 x (1/2 - 1/4 + 1/4 - 1/6 + 1/6-1/8 + ... + 1/98 - 1/100)
= 1/2 x (1/2 - 1/100)
= 1/2 x 98/100
= 98/200
ĐS: 98/200
LK
1
6 tháng 9 2021
\(\frac{13x-2}{2x+5}=-\frac{27}{5-x}\)ĐK : \(x\ne-\frac{5}{2};5\)
\(\Rightarrow\left(13x-2\right)\left(5-x\right)=-27\left(2x+5\right)\)
\(\Leftrightarrow65x-13x^2-10+2x=-54x-135\)
\(\Leftrightarrow-13x^2+121x+125=0\Leftrightarrow x=10,24...;x=-0,93...\)
XO
23 tháng 6 2020
Ta có :A = \(\frac{5}{3.4}+\frac{5}{4.6}+\frac{5}{5.8}+...+\frac{5}{40.78}=\frac{5}{2.2.3}+\frac{5}{2.3.4}+\frac{5}{2.4.5}+...+\frac{5}{2.39.40}\)
\(=\frac{5}{2}\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{39.40}\right)=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{39}-\frac{1}{40}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{40}\right)=\frac{5}{2}.\frac{19}{40}=\frac{19}{16}\)
5 tháng 8 2016
\(\frac{-5}{3}-\left(\frac{4}{5}-\frac{1}{2}\right)-\left|\frac{3}{4}-\frac{5}{2}+\frac{1}{3}\right|\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\left|\frac{3}{4}+\frac{-5}{2}+\frac{1}{3}\right|\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\left(\frac{3}{4}+\frac{-5}{2}+\frac{1}{3}\right)\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\frac{3}{4}+\frac{5}{2}-\frac{1}{3}\)
\(=\left(\frac{-5}{3}-\frac{1}{3}\right)+\left(\frac{1}{2}+\frac{5}{2}\right)-\left(\frac{4}{5}+\frac{3}{4}\right)\)
\(=\frac{-6}{3}+\frac{6}{2}-\left(\frac{16}{20}+\frac{15}{20}\right)\)
\(=-2+3-\frac{31}{20}\)
\(=1-\frac{31}{20}=\frac{-11}{20}\)
6 tháng 8 2016
\(\frac{-5}{3}-\left(\frac{4}{5}-\frac{1}{2}\right)-\left|\frac{3}{4}-\frac{5}{2}+\frac{1}{3}\right|\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\left|\frac{3}{4}+\frac{-5}{2}+\frac{1}{3}\right|\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\left(\frac{3}{4}+\frac{-5}{2}+\frac{1}{3}\right)\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\frac{3}{4}+\frac{5}{2}-\frac{1}{3}\)
\(=\left(\frac{-5}{3}-\frac{1}{3}\right)+\left(\frac{1}{2}+\frac{5}{2}\right)-\left(\frac{4}{5}+\frac{3}{4}\right)\)
\(=\frac{-6}{3}+\frac{6}{2}-\left(\frac{16}{20}+\frac{15}{20}\right)\)
\(=\frac{-6}{3}+\frac{6}{2}-\left(\frac{16}{20}+\frac{15}{20}\right)\)
\(=1-\frac{31}{20}=\frac{-11}{20}\)
28 tháng 8 2018
|x - 5| = |-7|
=> |x - 5| = 7
=> \(\orbr{\begin{cases}x-5=7\\x-5=-7\end{cases}}\)
=> \(\orbr{\begin{cases}x=12\\x=-2\end{cases}}\)
28 tháng 8 2018
c ) | x - 5 | = | -7 |
cs 2 trường hợp
TH1 x-5=7
x=5+7
x=12
TH2 x-5=-7
x=-7+5
x=-2
# chi kute@#
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
26 tháng 6 2023
A = \(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+\(\dfrac{5}{11.16}\)+\(\dfrac{5}{16.21}\)+...+\(\dfrac{5}{101.106}\)
A = \(\dfrac{1}{1}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{106}\)
A = \(\dfrac{105}{106}\)
B = \(\dfrac{3}{1.4}\) +\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{97.100}\)
B = \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
B = \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\)
B = \(\dfrac{99}{100}\)
C = \(\dfrac{1}{2.7}+\dfrac{1}{7.12}\) + \(\dfrac{1}{12.17}\)+...+ \(\dfrac{1}{97.102}\)
C= \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{5}{2.7}+\dfrac{5}{7.12}+\dfrac{5}{12.17}+...+\dfrac{5}{97.102}\))
C = \(\dfrac{1}{5}\)\(\times\)(\(\dfrac{1}{2}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{17}\)+...+ \(\dfrac{1}{97}\) - \(\dfrac{1}{102}\))
C = \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{102}\))
C = \(\dfrac{1}{5}\) \(\times\) \(\dfrac{25}{51}\)
C = \(\dfrac{5}{51}\)
D = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)
D = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+\(\dfrac{1}{7.8}\)+ \(\dfrac{1}{8.9}\)
D = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)
D = \(\dfrac{1}{1}\) - \(\dfrac{1}{9}\)
D = \(\dfrac{8}{9}\)
E = \(\dfrac{3}{2.4}\)+\(\dfrac{3}{4.6}\)+\(\dfrac{3}{6.8}\)+...+\(\dfrac{3}{98.100}\)
E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\)+ \(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{98.100}\))
E = \(\dfrac{3}{2}\)\(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{98}\) - \(\dfrac{1}{100}\))
E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\))
E = \(\dfrac{3}{2}\) \(\times\) \(\dfrac{49}{100}\)
E = \(\dfrac{147}{200}\)
A=\(\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)
A=\(\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(A=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(A=\frac{5}{2}.\frac{49}{50}\)
\(A=\frac{49}{20}\)