\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1998}=3\)

\(\Leftrightarrow\left(\frac{x-1}{2006}-1\right)+\left(\frac{x-10}{1997}-1\right)+\left(\frac{x-19}{1998}-1\right)=0\)

\(\Leftrightarrow\frac{x-2007}{2006}+\frac{x-2007}{1997}+\frac{x-2007}{1998}=0\)

\(\Leftrightarrow\left(x-2007\right)\left(\frac{1}{2006}+\frac{1}{1997}+\frac{1}{1988}\right)=0\)

Dễ thấy cái đằng sau luôn > 0 nên x-2007=0 <=> x=2007

\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}=3\)

\(\Leftrightarrow\frac{x-2007}{2006}+\frac{x-2007}{1997}+\frac{x-2007}{1988}=0\)

\(\Leftrightarrow x=2007\)

✰ ღ๖ۣۜDαɾƙ ๖ۣۜBαηɠ ๖ۣۜSĭℓεηтღ✰

lắm tắt thế này đi thi ko đc điểm đâu nhóc =))

\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1998}=3\)

\(\Leftrightarrow\left(\frac{x-1}{2006}-1\right)+\left(\frac{x-10}{1997}-1\right)+\left(\frac{x-19}{1998}-1\right)=0\)

\(\Leftrightarrow\frac{x-2007}{2006}+\frac{x-2007}{1997}+\frac{x-2007}{1998}=0\)

\(\Leftrightarrow\left(x-2007\right)\left(\frac{1}{2006}+\frac{1}{1997}+\frac{1}{1998}\right)=0\)

Dễ thấy \(\left(\frac{1}{2006}+\frac{1}{1997}+\frac{1}{1998}\right)>0\)nên \(x-2007=0\Leftrightarrow x=2007\)

Đặt \(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}\left(1\right)\)

\(\left(1\right)\Leftrightarrow\frac{x-2007}{2006}=\frac{x-2007}{1997}=\frac{x-2007}{1998}=0\)

\(\Rightarrow x=2007\)

Em kiểm tra lại đề bài nhé! Thiếu đề rồi.

<=> (x-1/2006 - 1)+(x-10/1997 - 1)+(x-19/1988 - 1) = 0

<=> x-2007/2006 + x-2007/1997 + x-2007/1988 = 0

<=> (x-2007).(1/2006+1/1997+1/1988) = 0

<=> x-2007=0 ( vì 1/2006+1/1997+1/1988 > 0 )

<=> x=2007

Vậy x=2007

k mk nha

Bạn sửa đề lại nha.

\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}=3\)

=>\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}-3=0\)

=>\(\left(\frac{x-1}{2006}-1\right)+\left(\frac{x-10}{1997}-1\right)+\left(\frac{x-19}{1988}-1\right)=0\)

=>\(\frac{x-1-2006}{2006}+\frac{x-10-1997}{1997}+\frac{x-19-1988}{1988}=0\)

=>\(\frac{x-2007}{2006}+\frac{x-2007}{1997}+\frac{x-2007}{1988}=0\)

=>\(\left(x-2007\right).\left(\frac{1}{2006}+\frac{1}{1997}+\frac{1}{1988}\right)=0\)

Vì \(\frac{1}{2006}+\frac{1}{1997}+\frac{1}{1988}\ne0\)

=>x-2007=0

=>x=2007

\(\frac{x-1}{2016}+\frac{x-10}{1997}+\frac{x-19}{1988}=3\)

\(pt\Leftrightarrow\frac{x-1}{2016}-1+\frac{x-10}{1997}-1+\frac{x-19}{1988}-1=0\)

\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{1997}+\frac{x-2017}{1988}=0\)

\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{1997}+\frac{1}{1988}\right)=0\)

Dễ thấy: \(\frac{1}{2016}+\frac{1}{1997}+\frac{1}{1988}\ne0\)

\(\Rightarrow x-2017=0\Rightarrow x=2017\)

Sửa đề :

\(\dfrac{x+1}{2006}+\dfrac{x+10}{1997}+\dfrac{x+19}{1988}=-3\)

\(\Leftrightarrow\left(\dfrac{x+1}{2006}+1\right)+\left(\dfrac{x+10}{1997}+1\right)+\left(\dfrac{x+19}{1988}+1\right)=0\)

\(\Leftrightarrow\dfrac{x+2007}{2006}+\dfrac{x+2007}{1997}+\dfrac{x+2007}{1988}=0\)

\(\Leftrightarrow\left(x+1007\right)\left(\dfrac{1}{2006}+\dfrac{1}{1997}+\dfrac{1}{1988}\right)=0\)

Mà \(\dfrac{1}{2006}+\dfrac{1}{1997}+\dfrac{1}{1988}\ne0\)

\(\Leftrightarrow x+2007=0\)

\(\Leftrightarrow x=-2007\)

Vậy..

\(x-\frac{x}{3}=\frac{3}{57}:\frac{12}{19}\)

\(x-\frac{x}{3}=\frac{3}{57}\times\frac{19}{12}\)

\(x-\frac{x}{3}=\frac{1}{12}\)

\(\Rightarrow12x-4x=1\)

\(\Rightarrow8x=1\)

\(\Rightarrow x=\frac{1}{8}\)