\(A=5x^2+2y^2+2xy-26x-16y+54\) \(=2\left(y^2+y\left(x-8\right)+\dfrac{\left(x-8\right)^2}{2}\right)-\dfrac{\left(x-8\right)^2}{2}+5x^2-26x+54\)
\(=2\left(y+\dfrac{x-8}{2}\right)^2+\dfrac{9}{2}x^2-18x+22\)
\(=2\left(y+\dfrac{x-8}{2}\right)^2+\dfrac{9}{2}\left(x-2\right)^2+4\ge4\)
Dấu '' = '' xảy ra khi: \(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+\dfrac{x-8}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Vậy: Min A = 4 tại \(x=2;y=3.\)
\(A=5x^2+2y^2+2xy-26x-16y+54\)
\(2A=4y^2+10x^2+4xy-52x-32y+108\)
\(2A=4y^2+4xy-32y-52x+10x^2+108\)
\(2A=\left(2y\right)^2+4y\left(x-8\right)+x^2-16x+64+9x^2-36x+44\)
\(2A=\left(2y\right)^2+2.2y\left(x-8\right)+\left(x-8\right)^2+\)\(9\left(x^2-4x+4\right)+8\)
\(2A=\left(2y+x-8\right)^2+9\left(x-2\right)^2+8\ge8\)
\(=>A\ge4\)
Để A nhỏ nhất thì \(x-2=0=>x=2;2y+x-8=0< =>2y-6=0=>y=3\)
Vậy ..................
\(\)
câu B tương tự thui bạn:)))
\(B=\left(x+2y\right)^2+\left(x-4\right)^2+\left(y-1\right)^2-27\)
\(=2x^2+4xy+5y^2-8x-2y-10\)
\(=2\left(x^2+2x\left(y-2\right)+\left(y-2\right)^2\right)-2\left(y-2\right)^2+5y^2-2y+10\)
\(=2\left(x+\left(y-2\right)\right)^2+3y^2+6y-18\)
\(=2\left(x+y-2\right)^2+3\left(y+1\right)^2-21\ge-21\)
Dấu '' = '' xảy ra khi \(\Leftrightarrow\left\{{}\begin{matrix}y+1=0\\x+y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)
Vậy: Min B = -21 tại \(x=3;y=-1.\)
