a) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Rightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

Mà \(\left(\frac{1}{9}< \frac{1}{8}< \frac{1}{7}< \frac{1}{6}\right)\)nên \(\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)< 0\)

\(\Rightarrow x+10=0\Rightarrow x=-10\)

Vậy x = -10

b) \(\frac{x}{2012}+\frac{x+1}{2013}+\frac{x+2}{2014}+\frac{x+3}{2015}+\frac{x+4}{2016}=5\)

\(\Rightarrow\frac{x}{2012}-1+\frac{x+1}{2013}-1+\frac{x+2}{2014}-1\)

\(+\frac{x+3}{2015}-1+\frac{x+4}{2016}-1=0\)

\(\Rightarrow\frac{x-2012}{2012}+\frac{x-2012}{2013}+\frac{x-2012}{2014}\)\(+\frac{x-2012}{2015}+\frac{x-2012}{2016}=0\)

\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\)

Mà \(\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)nên x - 2012 = 0

Vậy x = 2012

a, (x+1)/9 +1 + (x+2)/8  =  (x+3)/7 + 1 + (x+4)/6 + 1

<=> (x+10)/9 +(x+10)/8 = (x+10)/7 + (x+10)/6

<=> (x+10). (1/9 +1/8 - 1/7 -1/6) =0

vì 1/9 +1/8 -1/7 - 1/6 khác 0

=> x+10=0

=> x=-10

B = 2 - 4 - 6 + 8 + 10 - 12 -14 + 16 + ...+ 2010 - 2012 - 2014 + 2016

   = (2 - 4 - 6 + 8 ) + ( 10 - 12 -14 +16 ) + ...+ ( 2010 - 2012 - 2014 + 2016 )

   = 0 + 0 +...+ 0 + 0 (có 252 số hạng 0)

    = 0

Ta có:  Từ 2 đến 2016 có \(\frac{2016-2}{2}+1=1008\)

B=(2-4)-(6-8)+(10-12)-(14-16)+...+(2010-2012)-(2014-2016)  như vậy có 1008:2=504 nhóm

=-2+2-2+2+...-2+2 có 504 số hạng trong đó có: 525 số -2 và 525 số +2

=0

\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{9.10.11}\)

\(\Rightarrow2.C=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{9.10.11}=\frac{1}{2}\left(\frac{2}{1.3}\right)+\frac{1}{3}\left(\frac{2}{2.4}\right)+\frac{1}{4}\left(\frac{2}{3.5}\right)+\)\(...+\frac{1}{10}\left(\frac{2}{9.11}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}\right)+\frac{1}{3}\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{4}\left(\frac{1}{3}-\frac{1}{5}\right)+...+\frac{1}{10}\left(\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{1}{2}-\frac{1}{2.3}+\frac{1}{3.2}-\frac{1}{3.4}+\frac{1}{4.3}-\frac{1}{4.5}+...+\frac{1}{10.9}-\frac{1}{10.11}\)

\(=\frac{1}{2}-\frac{1}{10.11}=\frac{27}{55}\)

=> C=27/110

Xét tử: \(2015+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)

\(=\left(1+1+...+1\right)+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)( trong ngoặc có 2015 số 1 )

\(=\left(1+\frac{2014}{2}\right)+\left(1+\frac{2013}{3}\right)+...+\left(1+\frac{1}{2015}\right)+1\)

\(=\frac{2016}{2}+\frac{2016}{3}+\frac{2016}{4}+...+\frac{2016}{2015}+\frac{2016}{2016}\)

\(=2016\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)\)

Ghép tử và mẫu  \(\frac{2016\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}=2016\)

Vậy \(A=2016\)

A = 2016

cộng 1 vào mỗi tỉ số ta được:

\(\frac{x+1}{2016}+1+\frac{x+2}{2015}+1+\frac{x+3}{2014}+1=\frac{x+4}{2013}+1+\frac{x+5}{2012}+\frac{x+6}{2011}\)

=>\(\frac{x+1}{2016}+\frac{2016}{2016}+\frac{x+2}{2015}+\frac{2015}{2015}+\frac{x+3}{2014}+\frac{2014}{2014}=\frac{x+4}{2013}+\frac{2013}{2013}+\frac{x+5}{2012}+\frac{2012}{2012}+\frac{x+6}{2011}+\frac{2011}{2011}\)

=>

\(\frac{x+2017}{2016}+\frac{x+2017}{2015}+\frac{x+2017}{2014}=\frac{x+2017}{2013}+\frac{x+2017}{2012}+\frac{x+2017}{2011}\)

=>

\(\frac{x+2017}{2016}+\frac{x+2017}{2015}+\frac{x+2017}{2014}-\left(\frac{x+2017}{2013}+\frac{x+2017}{2012}+\frac{x+2017}{2011}\right)=0\)

=>

\(\frac{x+2017}{2016}+\frac{x+2017}{2015}+\frac{x+2017}{2014}-\frac{x+2017}{2013}-\frac{x+2017}{2012}-\frac{x+2017}{2011}=0\)

=>(x+2017).(1/1016+1/2015+1/2014-1/2013-1/2012-1/2011)=0

dễ thấy 1/2016<1/2015<1/2014<1/2013<1/2012<1/2011

=>1/2016+...-1/2011 khác 0

=>x+2017=0

=>x=-2017

nhớ tick

\(TA-CO':\)

\(A=\frac{4+\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}}{7+\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}}\)

\(A=\frac{4\left(\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2012}-\frac{1}{2013}\right)}{7\left(\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2012}-\frac{1}{2013}\right)}\)

\(A=\frac{4}{7}\)

\(B=\frac{1+2+...+2^{2013}}{2^{2015}-2}\)

ĐẶT \(C=1+2+...+2^{2013}\)

\(\Rightarrow2C=2+2^2+...+2^{2014}\)

\(\Rightarrow2C-C=\left(2+2^2+...+2^{2014}\right)-\left(1+2+...+2^{2013}\right)\)

\(\Rightarrow C=2^{2014}-2\)

\(\Rightarrow B=\frac{2^{2014}-1}{2^{2015}-2}\)

\(B=\frac{2^{2014}-1}{2\left(2^{2014}-1\right)}\)

\(B=\frac{1}{2}\)

\(\Rightarrow A-B=\frac{3}{7}-\frac{1}{2}=\frac{6}{14}-\frac{7}{14}\)

\(A-B=\frac{6-7}{14}=\frac{-1}{14}\)

VẬY, \(A-B=\frac{-1}{14}\)

Đặt phân thức trên là D

=> D=(1+1+1+1+...+1+2013/2+2012/3+...+2/2013+1/2014)/(1/2+1/3+1/4+...+1/2014)

=> D=(1+2013/2+1+2012/3+1+2011/4+...+1+2/2013+1+1/2014+1)/(1/2+1/3+1/4+1/5+...+1/2014)

=> D=(2015/2+2015/3+2015/4+...+2015/2013+2015/2014+1)/(1/2+1/3+1/4+...+1/2014)

=> D=[2015*(1/2+1/3+1/4+1/5+....+1/2014)]/(1/2+1/3+1/4+1/5+...+1/2014)

=> D=2015

UwU

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đùa thôi đáp án: 2015 nha bn

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d= 2015