Question

[Phan Van Khanh]

(2009-X)² +(2009-X)(X-2010)+(X-2010)²

(2009-X)²-(2009-X)(X-2010)+(X-2010)²

=19/49

mong cac ban thong cam minh chua quen dau gach la phan so

Answer 1

[Phước Nguyễn]

\(\frac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\frac{19}{49}\)

\(ĐKXĐ:\)  \(x\ne2009\)   \(;\)  \(x\ne2010\)

Đặt  \(a=x-2010\)  (với  \(a\ne0\)  ), ta được:

\(\frac{\left(a+1\right)^2-\left(a+1\right)a+a^2}{\left(a+1\right)^2+\left(a+1\right)a+a^2}=\frac{19}{49}\)  \(\Leftrightarrow\)  \(\frac{a^2+a+1}{3a^2+3a+1}=\frac{19}{49}\)  \(\Leftrightarrow\)  \(49a^2+49a+49=57a^2+57a+19\)

\(\Leftrightarrow\)  \(8a^2+8a-30=0\)  \(\Leftrightarrow\)  \(4a^2+4a-15=0\)  \(\Leftrightarrow\)  \(\left(2a+1\right)^2-16=0\)

\(\Leftrightarrow\)  \(\left(2a-3\right)\left(2a+5\right)=0\)  \(\Leftrightarrow\)  \(^{a=\frac{3}{2}}_{a=-\frac{5}{2}}\)  ( thỏa mãn điều kiện )

Do đó,  \(x=\frac{4023}{2}\)  hoặc \(x=\frac{4015}{2}\)  (thỏa mãn \(ĐKXĐ\) )

Vậy,  \(S=\left\{\frac{4023}{2};\frac{4015}{2}\right\}\)