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

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

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

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

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.

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

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

=>x-2007=0

=>x=2007

Bỏ 1/3 ở cuối nhé

Ta có:(1+1999/2)+(1+1998/3)+...(2/1999)(có 1998 tổng<=>1998 số 1)+(2000 - 1998)+400

        = 2001/2+2001/3+...+2001/1999+402

        =2001.(1/2+1/3+...+1/1999)+402(1)

      Thay (1) vào biểu thức trên và tính(tự tính nha!,tk cho mk!!!)

\(\frac{x+15}{2000}+\frac{x+16}{1999}=\frac{x+17}{1998}+\frac{x+18}{1997}\)

\(\Leftrightarrow\frac{x+15}{2000}+1+\frac{x+16}{1999}+1=\frac{x+17}{1998}+1+\frac{x+18}{1997}+1\)

\(\Leftrightarrow\frac{x+2015}{2000}+\frac{x+2015}{1999}=\frac{x+2015}{1998}+\frac{x+2015}{1997}\)

\(\Leftrightarrow\frac{x+2015}{2000}+\frac{x+2015}{1999}-\frac{x+2015}{1998}-\frac{x+2015}{1997}=0\)

\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2000}+\frac{1}{1999}-\frac{1}{1998}-\frac{1}{1997}\right)=0\)

Có: \(\frac{1}{2000}+\frac{1}{1999}-\frac{1}{1998}-\frac{1}{1997}\ne0\)

\(\Rightarrow x+2015=0\Rightarrow x=-2015\)

nhớ là có ai làm rồi mà quên =))

khá dài à =)) bấm máy mỏi :v

<=> (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