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Nguyen Thuy Tien
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Quang Anh Đặng
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Trần Nhật Dương
22 tháng 7 2018 lúc 16:53

\(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{2013}\)

\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)

\(\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)

\(1-\frac{1}{x+1}=2013\)

\(\frac{x}{x+1}=2013\)

x = 2013x + 2013

Vậy ko có gt của x

Lê Việt Hùng
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nguyễn văn kiệt
15 tháng 4 2017 lúc 22:07

Ta có: \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=2013+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)

\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=1+\left(1+\frac{2012}{2}\right)+...+\left(1+\frac{2}{2012}\right)+\left(1+\frac{1}{2013}\right)\)

\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=\frac{2014}{2014}+\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2012}+\frac{2014}{2013}\)

\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=2014.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}\right)\)

\(\Rightarrow x=2014\)

Lưu ý: số 2013 ở dòng T2 được tách ra làm 2013 số 1

cô gái lạnh lùng
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Luu Phan Hai Dang
21 tháng 3 2019 lúc 22:00

 b,\(\Rightarrow\)\(\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right):2=\frac{2013}{2015}:2\)

\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2013}{4030}\)

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2013}{4030}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{4030}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2013}{4030}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2015}\)

\(\Rightarrow\)\(x+1=2015\)

\(\Rightarrow x=2014\)

Nguyễn Hùng Cường
21 tháng 3 2019 lúc 22:06

a, 2/3x -3/2.x-1/2x=5/12

    x.(2/3-3/2-1/2)=5/12

                 x. -4/3=5/12

                          x=5/12:-4/3

                          x=-5/16

b,2/6+2/12+2/20+...+2/x.(x+1)=2013/2015

   2/2.3+2/3.4+2/4.5+...+2/x.(x+1)=2013/2015

   1/2(1-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1)=2013/2015

                                                1/2(1-1/x+1)=2013/2015

                                                 1-1/x+1=2013/2015 : 1/2

                                                  1-1/x+1=4206/2015

                                                      suy ra đề sai

                                                

                                                       

Người Con Của Rồng
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phạm nguyễn phương chi
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✓ ℍɠŞ_ŦƦùM $₦G ✓
1 tháng 6 2018 lúc 8:09

Cái này lớp 6 : 

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+......+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)

\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

<=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{2}{4026}=\frac{1}{2013}\)

\(\Leftrightarrow x+1=2013\)

=> x = 2012

Arima Kousei
1 tháng 6 2018 lúc 8:10

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

\(\Rightarrow1-\frac{2}{x+1}=\frac{2011}{2013}\)

\(\Rightarrow\frac{2}{x+1}=1-\frac{2011}{2013}\)

\(\Rightarrow\frac{2}{x+1}=\frac{2}{2013}\)

\(\Rightarrow x+1=2013\)

\(\Rightarrow x=2013-1\)

\(\Rightarrow x=2012\)

Vậy \(x=2012\)

~ Ủng hộ nhé 

Cristiano Ronaldo
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Thanh Tùng DZ
11 tháng 12 2017 lúc 21:17

\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=\frac{2013}{1}+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)

\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=\left(\frac{2012}{2}+1\right)+...+\left(\frac{2}{2012}+1\right)+\left(\frac{1}{2013}+1\right)+1\)

\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=\frac{2014}{2}+...+\frac{2014}{2012}+\frac{2014}{2013}+\frac{2014}{2014}\)

\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=2014.\left(\frac{1}{2}+...+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\right)\)

\(x=\frac{2014.\left(\frac{1}{2}+...+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)

\(x=2014\)

Vũ Thị Thảo Quyên
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Nguyễn Khánh Linh
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Xyz OLM
23 tháng 8 2019 lúc 14:47

1) Tính : 

a) \(\left(2008.2009.2010.2011\right).\left(1+\frac{1}{2}:\frac{2}{3}-\frac{4}{3}\right)\)

\(=\left(2008.2009.2010.2011\right).\left(1+\frac{1}{3}-\frac{4}{3}\right)\)

\(=\left(2008.2009.2010.2011\right).\left(\frac{4}{3}-\frac{4}{3}\right)\)

\(=\left(2008.2009.2010.2011\right).0\)

\(=0\)

2) Tìm x 

a) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{2013}:2\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2013}\)

\(\Rightarrow x+1=2013\)

\(\Rightarrow x=2012\)

b) \(\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}.\frac{1}{6}.\left(x-1,010\right)=\frac{1}{360}-\frac{1}{720}\)

\(\Rightarrow\frac{1}{2.3.4.5.6}.\left(x-1,01\right)=\frac{1}{720}\)

\(\Rightarrow\frac{1}{720}.\left(x-1,01\right)=\frac{1}{720}\)

\(\Rightarrow x-1,01=\frac{1}{720}:\frac{1}{720}\)

\(\Rightarrow x-1,01=1\)

\(\Rightarrow x=1+1,01\)

\(\Rightarrow x=2,01\)