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7 tháng 5 2016

có x mà ko có VP

7 tháng 5 2016
1/2 + 1/6 + 1/12 + 1/20 + ... + 1/x(x + 1) = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/x(x+1) = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1 = 1 - 1/x+1 = x+1/x+1 - 1/x+1 = x/x+1
27 tháng 4 2018

\(\frac{1}{2}\)+\(\frac{1}{6}\)+\(\frac{1}{12}\)+....+\(\frac{1}{x\left(x+1\right)}\)=\(\frac{2015}{2016}\)

\(\frac{1}{1x2}\)+\(\frac{1}{2x3}\)+\(\frac{1}{3x4}\)+.....+\(\frac{1}{x\left(x+1\right)}\)=\(\frac{2015}{2016}\)

\(1\)-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-.....+\(\frac{1}{x}\)-\(\frac{1}{x+1}\)=\(\frac{2015}{2016}\)

1-\(\frac{1}{x+1}\)                                                                     = \(\frac{2015}{2016}\)

\(\frac{1}{x+1}\)                                                                         =1- \(\frac{2015}{2016}\)

\(\frac{1}{x+1}\)                                                                          = \(\frac{1}{2016}\)

\(\Rightarrow\)x + 1= 2016

\(\Rightarrow\)x = 2015

5 tháng 5 2016

\(\frac{1}{2}+\frac{1}{6}+.....+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

\(=1-\frac{1}{x+1}=\frac{2015}{2016}\)

\(=\frac{1}{x+1}=\frac{1}{2016}\)

=> x + 1 = 2016

=> x =2015

5 tháng 5 2016

=>1- 1/2 + 1/2 - 1/3+.....+1/x - 1/(x+1) = 2008/2009

=>1 - 1/(x+1) = 2008/2009

=>1 - 1/(x+1) =1-1/1009

=>1/(x+1)=1/2009

=>x+1=2009

=>x=2008.Vậy x=2008

7 tháng 5 2016

VẾ TRÁI LÀ:

A=1/2.3+1/3.2+1/4.5+...+1/x[x+1]

A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/n-1/n+1

A=1-1/n+1

1-1/n+1=2015/2016

1/n+1=1-2015/2016

1/n+1=1/2016

n=2016-1

n=2015

Vế trái là 

S=1/2+1/6+1/12+1/20+..+1/x(x+1)

S=1/1.2+1/2.3+1/3.4+1/4.5+...+1/x.(x+1)

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

S=1-1/(x+1)=Vế phải=2015/2016

<=>1-1/(x+1)=2015/2016

1/(x+1)=1/2016

=>x+1=2016

x=2015

Ủng hộ mk mha Chí Tiến

14 tháng 9 2021

b   fhhhggbbbvgftfhmnhmcvch

AH
Akai Haruma
Giáo viên
20 tháng 10 2024

Lời giải:

$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{x(x+1)}=\frac{3}{8}$

$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{x(x+1)}=\frac{3}{8}$

$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.....+\frac{1}{6}-\frac{1}{7}+\frac{1}{x(x+1)}=\frac{3}{8}$

$1-\frac{1}{7}+\frac{1}{x(x+1)}=\frac{3}{8}$
$\frac{6}{7}+\frac{1}{x(x+1)}=\frac{3}{8}$

$\frac{1}{x(x+1)}=\frac{3}{8}-\frac{6}{7}=\frac{-27}{56}$

Kết quả này không phù hợp lắm.

Bạn xem lại đề nhé. 

29 tháng 4 2015

Ta có:

(1/2 + 1/4 + 1/8 + 1/16) = 8/16 + 4/16 + 2/16 + 1/16 = 15/16.

1/2 + 1/6 + 1/12 + 1/20 +…+ 1/132 = 1/(1.2) + 1/(2.3) + 1/(3.4) + 1/(4.5) +…+1/(11.12)

= (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1/4) + (1/4 – 1/5) +…+ (1/11 – 1/12)

= 1 – 1/12 = 11/12

Vậy x = (15/16) : (11/12) = 45/44.

29 tháng 4 2015

(1/2+1/4+1/8+1/16):x=1/2+1/6+1/12+1/20+...+1/132

(1-1/2+1/2-1/4+1/4-1/8+1/8-1/16):x=1/1x2+1/2x3+1/3x4+1/4x5+...+1/11x12

(1-1/16):x=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/11-1/12

15/16:x=1-1/12

15/16:x=11/12

x=15/16:11/12

x=45/44

14 tháng 7 2017

Sửa đề : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{100}\)

\(\Leftrightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{100}\)

\(\Leftrightarrow\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+....+\frac{\left(x+1\right)-x}{x\left(x+1\right)}=\frac{98}{100}\)

\(\Leftrightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}=\frac{98}{100}\)

\(\Leftrightarrow\frac{1}{1}-\frac{1}{\left(x+1\right)}=\frac{98}{100}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{1}-\frac{98}{100}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{50}\)

\(\Leftrightarrow x=50-1=49\)

14 tháng 7 2017

Sửa đề: \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{100}\)

(=) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{100}\)

(=)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{98}{100}\)

(=)\(1-\frac{1}{x+1}=\frac{98}{100}\)

(=)\(\frac{1}{x+1}=1-\frac{98}{100}\)

(=)\(\frac{1}{x+1}=\frac{1}{50}\)=> \(x+1=50\)

                                    \(x=50-1\)

                                    \(x=49\)

T_i_c_k cho mình nha,thanks you so much!