1/2,nhân x,trừ 1/1.2,trừ1/2.3-....1/45.46=-2
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NT
1
1 tháng 8 2019
\(\frac{1}{2.x}-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{45.46}=-2\)
\(\frac{1}{2.x}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{45.46}\right)=-2\)
\(\frac{1}{2.x}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{45}-\frac{1}{46}\right)=-2\)
\(\frac{1}{2.x}-\left(1-\frac{1}{46}\right)\)
\(\frac{1}{2.x}-\frac{45}{46}=-2\)
\(\frac{1}{2.x}=-2+\frac{45}{46}\)
\(\frac{1}{2.x}=\frac{-47}{46}\)
\(2x=\frac{46}{-47}\)
\(x=\frac{46}{-47}:2=\frac{-23}{47}\)
1 tháng 8 2019
\(\frac{1}{2.x}-\frac{1}{1.2}-\frac{1}{2.3}-......-\frac{1}{45.46}=-2\)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: \(\frac{1}{2.x}-\frac{45}{46}=-2\)
\(\frac{1}{2.x}=\frac{-47}{46}\)
\(\frac{-47}{-94.x}=\frac{-47}{46}\)
\(\Rightarrow x=\frac{-23}{47}\)
TT
1
4 tháng 6 2020
\(75\%-1\frac{1}{2}+0,5:\frac{5}{12}-\frac{-1^2}{2}\)
\(=\frac{75}{100}-\frac{3}{2}+\frac{5}{10}.\frac{12}{5}-\frac{1}{2}\)
\(=\frac{3}{4}-\frac{3}{2}+\frac{6}{5}-\frac{1}{2}\)
\(=\frac{15}{20}-\frac{30}{20}+\frac{24}{20}-\frac{10}{20}\)
\(=\frac{15-30+24-10}{20}\)
\(=\frac{-1}{20}\).
NN
0
21 tháng 7 2023
\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(100-10\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)
\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=1.2=2\)
\(\Rightarrow\left(x+\dfrac{206}{100}\right)=\dfrac{5}{2}:2=\dfrac{5}{2}.\dfrac{1}{2}=\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{5}{4}-\dfrac{206}{100}=\dfrac{125}{100}-\dfrac{206}{100}\)
\(\Rightarrow x=-\dfrac{81}{100}\)
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?
\(=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}\)
BN
1
9
30 tháng 1 2023
Tham khảo:
A=1.2+2.3+3.4+...+2013.2014
3A = 1.2.3 + 2.3.3 + 3.4.3 +...+ 2013.2014.3
Mà: 1.2.3 = 1.2.3
2.3.3 = 2.3.4 - 2.3.1
3.4.3 = 3.4.5 - 3.4.2
2012.2013.3 = 2012.2013.2014 - 2012.2013.2011
2013.2014.3 = 2013.2014.2015 - 2013.2014.2012
=> 3S = 2013.2014.2015
=> A = 2013.2014.2015 / 3 = 2723058910