\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\)( * )
Đặt \(a=2x^2+x-2013\)
\(\)Đặt \(b=x^2-5x-2012\)
Khi đó ( * ) trở thành:
\(a^2+4b^2=4ab\)
\(\Leftrightarrow a^2+4b^2-4ab=0\)
\(\Leftrightarrow a^2-4ab+4b^2=0\)
\(\Leftrightarrow\left(a-2b\right)^2=0\)
\(\Leftrightarrow a-2b=0\)
\(\Leftrightarrow a=2b\)
\(\Leftrightarrow2x^2+x-2013=2\left(x^2-5x-2012\right)\)
\(\Leftrightarrow2x^2+x-2013-2x^2+10x+4024=0\)
\(\Leftrightarrow11x+2011=0\)
\(\Leftrightarrow x=\dfrac{-2011}{11}\)
Vậy...
đặt: \(x=2x^2+x-2013\\ y=x^2-5x-2012\), khi đó:
\(x^2+4y^2=4xy\\ \Leftrightarrow x^2-4xy+y^2=0\\ \Leftrightarrow\left(x-2y\right)^2=0\Rightarrow x-2y=0\\ \Leftrightarrow x=2y\\ \Rightarrow2x^2+x-2013=2x^2-10x-4024\)
\(\Leftrightarrow11x=-2011\\ \Leftrightarrow x=-\dfrac{2011}{11}\)
vậy ........
\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\)
Đặt \(\left\{{}\begin{matrix}2x^2+x-2013=a\\x^2-5x-2012=b\end{matrix}\right.\) thì ta có:
\(a^2+4b^2=4ab\)\(\Rightarrow a^2+4b^2-4ab=0\)
\(\Rightarrow\left(a-2b\right)^2=0\Rightarrow a-2b\Rightarrow a=2b\)
Tức là \(2x^2+x-2013=2\left(x^2-5x-2012\right)\)
\(\Leftrightarrow2x^2+x-2013=2x^2-10x-4024\)
\(\Leftrightarrow11x+2011=0\Leftrightarrow11x=-2011\Rightarrow x=-\dfrac{2011}{11}\)
