Đặt \(A=10x^2+15x+8x-12xy+4y^2\)
\(=\left(9x^2-12xy+4y^2\right)+\left(x^2+15x+8x\right)\)
\(=\left(3x-2y\right)^2+\left(x^2+23x\right)\)
\(=\left(3x-2y\right)^2+\left(x^2+2.x\dfrac{23}{2}+\dfrac{529}{4}-\dfrac{529}{4}\right)\)
\(=\left(3x-2y\right)^2+\left(x+\dfrac{23}{2}\right)^2-\dfrac{529}{4}\)
Vì \(\left(3x-2y\right)^2\ge0\) với mọi x,y
\(\left(x+\dfrac{23}{2}\right)^2\ge0\) với mọi x
\(\Rightarrow\left(3x-2y\right)^2+\left(x+\dfrac{23}{2}\right)^2-\dfrac{529}{4}\ge-\dfrac{529}{4}\) với mọi x,y
\(\Rightarrow Amin=-\dfrac{529}{4}\Leftrightarrow\left\{{}\begin{matrix}3x-2y=0\\x+\dfrac{23}{2}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x=2y\\x=-\dfrac{23}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3.\left(-\dfrac{23}{2}\right)=2y\\x=-\dfrac{23}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-\dfrac{69}{2}=2y\\x=-\dfrac{23}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=-\dfrac{69}{4}\\x=-\dfrac{23}{2}\end{matrix}\right.\)
Vậy giá trị nhỏ nhất của biểu thức là -529/4 khi y = -69/4 và x = -23/2