1 ) Xét : \(x^2-9=0\)
\(\Leftrightarrow x^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy nghiệm của đ/t trên là : \(\left[{}\begin{matrix}3\\-3\end{matrix}\right.\)
2 ) \(2\left(x-y\right)\left(x-y\right)+\left(2x-y\right)^2-\left(x-y\right)^2\)
\(=2\left(x-y\right)^2+\left(2x-y\right)^2-\left(x-y\right)^2\)
\(=\left(x-y\right)^2+\left(2x-y\right)^2\)
\(=x^2-2xy+y^2+4x^2-4xy+y^2\)
\(=5x^2-6xy+2y^2\)
3 ) \(x-x^2-3=-\left(x^2-x+3\right)=-\left(x^2-x+\dfrac{1}{4}+\dfrac{11}{4}\right)=-\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}\right]=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\forall x\)Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
Vậy Max của b/t trên là : \(-\dfrac{11}{4}\Leftrightarrow x=\dfrac{1}{2}\)
