<=> (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

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

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

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

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

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 \(x-2007=0\Leftrightarrow x=2007\)

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

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

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

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

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

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

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.

\(a.\dfrac{x-2}{2000}+\dfrac{x-3}{1999}=\dfrac{x-4}{1998}+\dfrac{x-5}{1997}\\ \Leftrightarrow\dfrac{x-2}{2000}-1+\dfrac{x-3}{1999}-1=\dfrac{x-4}{1998}-1+\dfrac{x-5}{1997}-1\\ \Leftrightarrow\dfrac{x-2}{2000}-\dfrac{2000}{2000}+\dfrac{x-3}{1999}-\dfrac{1999}{1999}=\dfrac{x-4}{1998}-\dfrac{1998}{1998}+\dfrac{x-5}{1997}-\dfrac{1997}{1997}\\ \Leftrightarrow\dfrac{x-2002}{2000}+\dfrac{x-2002}{1999}=\dfrac{x-2002}{1998}+\dfrac{x-2002}{1997}\\ \Leftrightarrow\dfrac{x-2002}{2000}+\dfrac{x-2002}{1999}-\dfrac{x-2002}{1998}-\dfrac{x-2002}{1997}=0\\ \Leftrightarrow\left(x-2002\right)\left(\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\right)=0\\ \)

\(Do:\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\ne0\\ \Rightarrow x-2002=0\\ \Leftrightarrow x=2002\\ Vậy:S=\left\{2002\right\}\)

Mấy câu khác tương tự :v

b: \(\Leftrightarrow\left(\dfrac{148-x}{25}-1\right)+\left(\dfrac{169-x}{23}-2\right)+\left(\dfrac{186-x}{21}-3\right)+\left(\dfrac{199-x}{19}-4\right)=0\)

=>123-x=0

=>x=123

c: \(\Leftrightarrow\dfrac{x-2}{2017}+1=\dfrac{x-1}{2018}+\dfrac{x}{2019}\)

\(\Leftrightarrow\left(\dfrac{x-2}{2017}-1\right)=\left(\dfrac{x-1}{2018}-1\right)+\left(\dfrac{x}{2019}-1\right)\)

=>x-2019=0

=>x=2019

a: \(\Leftrightarrow\left(x-1\right)^2=81\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)