a) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)

<=> \(\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)-\left(\frac{x-3}{2007}-1\right)-\left(\frac{x-4}{2006}-1\right)=0\)

<=> \(\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)

<=> \(\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)

<=> x - 2010 = 0 Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\)

<=> x = 2010

\(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|=4\left(x-4\right)\)

Ta thấy : \(\left|x-1\right|\ge0;\left|x-2\right|\ge0;\left|x-3\right|\ge0\)

=> \(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\ge0\)

=> 4 ( x - 4 ) \(\ge0\). Mà 4 > 0 => \(x-4\ge0=>x\ge4\)hay

\(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|=4\left(x-4\right)=>x-1+x-2+x-3=4\left(x-4\right)\) => 3x - 6 = 4x - 16

=> -6+16 = 4x - 3x => x = 10

\(\frac{\left(2007-x\right)^2+\left(2007-x\right)\left(x-2008\right)+\left(x-2008\right)^2}{\left(2007-x\right)^2-\left(2007-x\right)\left(x-2008\right)+\left(x-2008\right)^2}=\frac{19}{49}\)

điểu kiện xác định x khác 2007 and x khác 2008

Đặt a=x-2008 ( a khác 0 ,) ta có hệ thức

\(\frac{\left(a+1\right)^2-\left(a+1\right)a+a^2}{\left(a+1\right)^2+\left(a+1\right)a+a^2}=\frac{19}{49}\)

=>\(\frac{a^2+a+1}{3a^2+3a+1}=\frac{19}{49}\)

=>\(49a^2+49a+49=57a^2+57a+19\)

=>\(8a^2+8a-30=0\)

=>\(\left(2a-1\right)^2-4^2=0=>\left(2a-3\right)\left(2a+5\right)=0\)

=>\(\orbr{\begin{cases}a=\frac{3}{2}\\a=-\frac{5}{2}\end{cases}}\)(Thỏa mãn điều kiện)

Tự thay a xong suy ra x nhá 

Mệt lắm r

bài khó thế 

\(???\)\(\frac{19}{29}ak\)

ko sao , bạn cx nhân chéo lên tương tự như cách làm của mình xong => ra a mà làm nha . Hihi ..^^

Đặt x -2006 = y 

pt <=>  \(\frac{y^2-y\left(y-1\right)+\left(y-1\right)^2}{y^2+y\left(y-1\right)+\left(y-1\right)^2}=\frac{19}{49}\)

<=> \(\frac{y^2-y^2+y+y^2-2y+1}{y^2+y^2-y+y^2-2y+1}=\frac{19}{49}\)

<=> \(\frac{y^2-y+1}{3y^2-3y+1}=\frac{19}{49}\)

<=> \(49y^2-49y+49=57y^2-57y+19\)

<=> \(8y^2-8y-30=0\)

<=> \(4y^2-4y+15=0\)

Giải tiếp nha 

đầu bài có sai k ạ???

de bai hinh nhu khong sai ban a

mk giải cho mà saI CÓ đc tiền k

2023 =))

c) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)

\(\Leftrightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)

\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)

\(\Leftrightarrow\left(x-2010\right).\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)

\(\Leftrightarrow x-2010=0\)

\(\Leftrightarrow x=0+2010\)

\(\Rightarrow x=2010\)

Vậy \(x=2010.\)

Mình chỉ làm câu c) thôi nhé.

Chúc bạn học tốt!

a: \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)

\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)

=>x+1=0

hay x=-1

b: \(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)

=>x-2010=0

hay x=2010

c: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)

=>x=15