Question

[Uchiha Sasuke]

a, Giải phương trình: \(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...........+\frac{x-2012}{1}=2012\)

Answer 1

[๖Fly༉Donutღღ]

Phương trình đã cho tương đương với :

\(\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+\frac{x-3}{2010}-1+...+\frac{x-2012}{1}-1+2012=2012\)

\(\Leftrightarrow\)\(\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)

\(\Leftrightarrow\)\(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\right)=0\)

Tìm x theo như toán lớp 6 nha

\(x-2013=0\)

\(\Leftrightarrow\)\(x=2013\)

Answer 2

[vũ tiền châu]

ta có pt 

<=>\(\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+...+\frac{x-2012}{1}-1=0\)

<=>\(\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)

<=>\(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{1}\right)=0\Leftrightarrow x-2013=0\Leftrightarrow x=2013\)

^_^