1/2 .x-1/1.2-1/2.3-..........-1/45.46=-2
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NT
0
VN
4 tháng 2 2018
Ta có: 1/1.2+1/2.3+1/3.4+...+1/x(x+1)=2/3
=> 1-1/2+1/2-1/3+1/3-1/4+...+1/x-1/x+1=2/3
=>1-1/x+1=2/3
=>1/x+1=1/3
=>3=x+1
=>x=2
4 tháng 2 2018
Ta có\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{3}\)
=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2}{3}\)
=>\(1-\frac{1}{x+1}=\frac{2}{3}\)
=>\(\frac{1}{x+1}=1-\frac{2}{3}\)
=>\(\frac{1}{x+1}=\frac{1}{3}\)
=>\(x+1=3\)
=>\(x=2\)
NH
10 tháng 3 2023
a)
`1/1-1/2`
`=2/2-1/2`
`=1/2`
b)
`1/(1*2)+1/(2*3)`
`=1/1-1/2+1/2-1/3`
`=1/1-1/3`
`=3/3-1/3`
`=2/3`
c)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}\\ =\dfrac{99}{100}\)
d)
\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+...+\dfrac{3}{99\cdot100}\) đề phải như thế này chứ nhỉ?
\(=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}\)
LP
1
20 tháng 8 2021
Ta có: \(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Leftrightarrow100\left(\dfrac{1}{1}-\dfrac{1}{10}\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]\cdot2=89\)
\(\Leftrightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow x+\dfrac{103}{50}=5\)
hay \(x=\dfrac{147}{50}\)
16 tháng 3 2022
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2021}{2022}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{2021}{2022}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2022}\)
=>x+1=2022
hay x=2021
NN
1.Tính
A= (1-1/22).(1-1/32)...(1-1/1002)
B= -1/1.2-1/2.3-1/3.4-...-1/100.101
C= 1.2+2.3+3.4+...+100.101
2
13 tháng 9 2016
Lời giải :
Đặt S=1.2+2.3+3.4+4.5+…+99.100+100.101
3S=1.2.3+2.3.3+3.4.3+4.5.3+…+99.100.3+100.101.3
=1.2(3−0)+2.3(4−1)+3.4(5−2)+4.5(6−3)+…+99.100(101−98)+100.101(102−99)
=0.1.2-1.2.3+1.2.3-2.3.4+...+99.100.101-100.101.102
=100.101.102
S=100.101.34=343400
12 tháng 10 2022
1.Tính
a) Ta có:
A=(1-1/22).(1-1/32)...(1-1/1002)
=>A=3/22.8/32.....9999/1002
=>A=(1.3/2.2).(2.4/3.3).....(99.101/100.100)
=>A=(1.2.3.....99/2.3.4.....100).(3.4.5.....101/2.3.4.....100)
=>A=1/100.101/2
=>A=101/200
b) Ta có:
B=-1/1.2-1/2.3-1/3.4-...-1/100.101
=>B=-(1/1.2+1/2.3+1/3.4+...+1/100.101)
=>B=-(1-1/2+1/2-1/3+1/3-1/4+...+1/100-1/101)
=>B=-(1-1/101)
=>B=-100/101
c) Ta có:
C=1.2+2.3+3.4+...+100.101
=>3C=1.2.3+2.3.3+3.4.3+...+100.101.3
=>3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+100.101.(102-99)
=>3C=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+...+100.101.102
=>3C=100.101.102
=>3C=1030200
=>C=343400
Chúc bạn hok tốt nhé >:)!!!!!
NT
1
HN
15 tháng 7 2019
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-........-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
\(\Leftrightarrow\frac{9}{10}.100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}\right]=89\)
\(\Leftrightarrow90-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}\right]=89\)
\(\Leftrightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}=90-89=1\)
\(\Leftrightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right)=1.\frac{1}{2}=\frac{1}{2}\)
\(\Leftrightarrow x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}\)
\(\Leftrightarrow x+\frac{103}{50}=\frac{5}{2}.2\)
\(\Leftrightarrow x+\frac{103}{50}=5\)
\(\Leftrightarrow x=5-\frac{103}{50}\)
\(\Leftrightarrow x=\frac{250}{50}-\frac{103}{50}\)
\(\Leftrightarrow x=\frac{147}{50}\)
\(\frac{1}{2.x}-\frac{1}{1.2}-\frac{1}{2.3}-......-\frac{1}{45.46}=-2\)2
\(\frac{1}{2.x}-\left(\frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{45.46}\right)=-2\)
Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{45.46}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{45}-\frac{1}{46}\)
\(A=1-\frac{1}{46}=\frac{45}{46}\)
Ta có: \(\frac{1}{2.x}-\frac{45}{46}=-2\)
\(\frac{1}{2.x}=\frac{-47}{46}\)
\(\frac{-47}{-94.x}=\frac{-47}{46}\)
\(\Rightarrow x=\frac{-23}{47}\)