1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1x1-3+2+2+2+2+2+2+2+2+2-2-2-2-2-2+2+2+2+2+2+2=
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HL
21 tháng 9 2020
a,1+1=.2.
2+2=4
3+2=5
4+4=8
5+4=9
b, 0+1=.1
1-1=.0
2-1=.1
3-2=..1.
4-1=3
c, 1x1=.1
3x1=..3.
3x2=6
1x4=.4.
5x2=.10
d, 2:1=1
4:2=..2.
3:1=.3
* (4-2):2=.1.
* (2+1):3=.1.
CM
21 tháng 9 2020
a,1+1=2 2+2=4 3+2=5 4+4=8 5+4=9
b,0+1=1 1-1=0 2-1=1 3-2=1 4-1=3
c,1x1=1 3x1=3 3x2=6 1x4=4 5x2=10
d,2:1=2 4:2=2 3:1=3
*(4-2):2=2:2=1
*(2+1):3=3:3=1
BH
1
26 tháng 7 2018
\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\cdot100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=89\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\cdot100-\left[\frac{5}{2}:\left(x+\frac{103}{50}\right)\right]\cdot2=89\)
\(\left(1-\frac{1}{10}\right)\cdot100-\frac{5}{2}:\left(x+\frac{103}{50}\right)\cdot2=89\)
\(\frac{9}{10}\cdot100-\frac{5}{2}\cdot2:\left(x+\frac{103}{50}\right)=89\)
\(90-5\cdot\left(x+\frac{103}{50}\right)=89\)
\(5\cdot\left(x+\frac{103}{50}\right)=1\)
\(x+\frac{103}{50}=\frac{1}{5}\)
\(x=-\frac{93}{50}\)
NN
4
20 tháng 6 2016
1 + 1 x 1 +1 = 1 + 1 +1 = 3
2 + 2 x 2 +2 = 2 + 4 + 2 = 8
3 + 3 - 3 + 3 = 6 - 3 + 3 = 6
4 - 4 + 4 - 4 = 0 + 4 - 4 = 0
5 x 5 - 5 x 5 = 25 - 25 =0
1 + 1 + 2 + 3 + 4 + 5 = 16
1 x 1 x 2 x 3 x 4 x 5 = 120
20 tháng 6 2016
1 + 1 x 1 + 1 = 3
2 + 2 x 2 + 2 = 8
3 + 3 x 3 + 3 = 15
4 - 4 + 4 - 4 = 0
5 x 5 - 5 x 5 = 0
1 + 1 + 2 + 3 + 4 + 5 =16
1 x 1 x 2 x 3 x 4 x 5 = 120
k nh
PN
0
3 tháng 2 2022
a) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
=\(1-\dfrac{1}{6}\)=\(\dfrac{5}{6}\)
b) \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
=\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
=\(\dfrac{1.2}{3.5.2}+\dfrac{1.2}{5.7.2}+\dfrac{1.2}{7.9.2}+\dfrac{1.2}{9.11.2}+\dfrac{1.2}{11.13.2}\)
=\(\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\).
=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)=\(\dfrac{1}{2}.\dfrac{10}{39}\)=\(\dfrac{5}{39}\).
c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
=\(1-\dfrac{1}{8}=\dfrac{7}{8}\).
d) \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\)
=\(\dfrac{2^4}{2^5}+\dfrac{2^3}{2^5}+\dfrac{2^2}{2^5}+\dfrac{2}{2^5}+\dfrac{1}{2^5}\)
=\(\dfrac{2^4+2^3+2^2+2+1}{2^5}\)=\(\dfrac{2^5-1}{2^5}=\dfrac{31}{32}\).
e) \(\dfrac{1}{7}+\dfrac{1}{7^2}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{100}}=\dfrac{7^{99}+7^{98}+7^{97}+...+7+1}{7^{100}}=\dfrac{\dfrac{7^{100}-1}{6}}{7^{100}}=\dfrac{7^{100}-1}{6.7^{100}}\)
KA
G=(1-1/2^2).(1-1/3^2).(1-1/4^2).......(1-1/100^2)
H=(1-1/2^2).(1-1/3^2).(1-1/4^2)........(1-1/2018^2)
0
LN
1
1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1x1-3+2+2+2+2+2+2+2+2+2-2-2-2-2-2+2+2+2+2+2+2= 33