Tìm x, biết (x+4)/2008+(x+3)/2009=(x+2)/2010+(x+1)/2011

1) a. Ta có:\(\frac{x+4}{2008}+\frac{x+3}{2009}=\frac{x+2}{2010}+\frac{x+1}{2011}\)

\(\Rightarrow\frac{x+4}{2008}+1+\frac{x+3}{2009}+1=\frac{x+2}{2010}+1+\frac{x+1}{2011}+1\)

\(\Rightarrow\frac{x+4+2008}{2008}+\frac{x+3+2009}{2009}=\frac{x+2+2010}{2010}+\frac{x+1+2011}{2011}\)

\(\Rightarrow\frac{x+2012}{2008}+\frac{x+2012}{2009}=\frac{x+2012}{2010}+\frac{x+2012}{2011}\)

\(\Rightarrow\left(x+2012\right)\left(\frac{1}{2008}+\frac{1}{2009}\right)=\left(x+2012\right)\left(\frac{1}{2010}+\frac{1}{2011}\right)\)

\(\Rightarrow\left(x+2012\right)\left(\frac{1}{2008}+\frac{1}{2009}\right)-\left(x+2012\right)\left(\frac{1}{2010}+\frac{1}{2011}\right)=0\)

\(\Rightarrow\left(x+2012\right)\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)

\(\Rightarrow x+2012=0\)

\(\Rightarrow x=-2012\)

Bài 2:

a.Ta có: \(\frac{x+2y}{18}=\frac{1+4y}{24}\)

\(\Rightarrow24x+48y=18+72y\)

\(\Rightarrow24x+48y-72y=18\)

\(\Rightarrow24x-24y=18\)

\(\Rightarrow24\left(x-y\right)=18\)

\(\Rightarrow x-y=\frac{3}{4}\)

\(\Rightarrow y=x-\frac{3}{4}\)

thay \(y=x-\frac{3}{4}\)vào \(\frac{1+4y}{24}=\frac{1+x+6y}{6x}\)ta được \(\frac{1+4\times\left(x-\frac{3}{4}\right)}{24}=\frac{1+x+6\times\left(x-\frac{3}{4}\right)}{6x}\)

giải ra ta được x=7

\(\Rightarrow y=7-\frac{3}{4}=\frac{25}{4}\)

b. Đẻ A mang giá trị nuyên

\(\Leftrightarrow9+3n⋮n-4\)

\(\Leftrightarrow3n-12+21⋮n-4\)

\(\Leftrightarrow3\left(n-4\right)+21⋮n-4\)

\(\Leftrightarrow21⋮n-4\)

\(\Leftrightarrow n-4\inƯ_{\left(21\right)}=\left\{\pm1;\pm3;\pm7;\pm21\right\}\)

Ta có bảng sau:

Vậy \(n\in\left\{5;4;7;1;11;-3;25;-17\right\}\)thì A là số nguyên.

Thay n vào A và tính giá trị